but if we use L Hospital rule and differentiate the numerator and denominator then we have 1/1-0 which equals 1 so by that method the limit should exist and should be equal to one
@@winners-r4z We can only apply the L'Hospital's rule if the direct substitution returns an indeterminate form, that means 0/0 or ±∞/±∞. As bprp said, this is *NOT* an undeterminate form. The limit is clearly ∞, but depending on the direction from which we approach 1, the sign of ∞ changes
Well, in both cases the function approaches 1/0. However, where x is less than 1 and greater than 0, the function is negative. Likewise, when x is greater than 1 or less than 0 (which is irrelevant to this question), the function is positive.
I did learn about that (albeit 25 years ago, oh god, and not with this guy’s notation) and my first reaction was that it doesn’t specify from which side it is in the problem.
@@JasperJanssen ' … did you watch the video?' Yes, I did. 'And no, there is no such thing as “the standard” Do you seriously not know about the non-one-sided limits?
This limit problem is a good illustration of why making even a rough sketch graph of the function in question can shed a lot of light. Using a graphing as a qualitative analytical tool is too often overlooked.
While i do agree that graphs are amazingly helpful,i believe more complex problems would be better suited for them,idk.. to me, the answer felt glaringly obvious from the start.
@@levaniandgiorgi2358 I largely agree. I looked at the statement and pretty much saw the answer immediately, but then I have the advantage over freshman students of having done math for close to 60 years. I’ve encouraged students to not shy away from sketching Bode plots or pole/zero diagrams in the EE courses I’ve taught. It is handy to look at a problem with more than one method to avoid mistakes. The backup method doesn’t need to be precise, just accurate enough to confirm your thinking is on track.
Graphs do what Graphs are supposed to do, give you a visual representation of the abstract equation It's helpful for people who are stronger visual learners to link the reasoning and the answer together
Graphs are very powerful. It's really hard to believe calculus was invented without using them. People make fun of 'trivial' stuff like rolle's theorem, but good luck proving them without graphs.
Note to self: "L'Hôpital's rule can only be applied in the case where direct substitution yields an indeterminate form, meaning 0/0 or ±∞/±∞. So if f and g are defined, L'Hôpital would be applicable only if the value of both f and g is 0. "
I used to love asking my students on these if their denominator was positive zero or negative zero. The transition from initial confusion to a-ha was one of my favorite gems from teaching.
@@krishnannarayanan8819 you could assume some number "h" which is a very small positive number. positive 0 means 0+h and negative 0 means 0-h . basically 0+ and 0- are approaching 0 from right and left sides respectively.
I just woke up and this video was suggested. I haven't touched in limits for almost a decade so my thought was "Ive forgetten all of this". I've opened the video,watched for 3 minutes and I could feel the knowledge coming back 😂 so weird
@@jamescollier3 I never really liked math as a whole in college. Then once I graduated, I started learning it in depth, on my own . Then I started loving it. Now it has been 7 years since I graduated and I still learn it
Knowing how this function's graph behaves gives all of the intuition needed. Vertical asymptote at x=1, positive to the right, negative between 0 and 1. My students often dive into the calculus without thinking about the precalculus. Sure it can be done without the precalc, but the confidence gets a big boost when we think about the graph first.
Approaching a limit doesn't require a Y-axis, you're needlessly complicating the concept and conditioning them with a euclidian bias in their thinking about numbers.
@No-cg9kj On any calculus exam, this question would be a multiple-choice or short answer question where you’re either right or wrong, no partial credit. Graphing is sometimes way faster than doing the math. just visualizing the graph i solved this problem in probably 2-3 seconds
@@iamcoolkinda 'On any calculus exam, this question would be a multiple-choice or short answer question where you’re either right or wrong, no partial credit' Literally none of the math exams that I took at university had multiple-choice questions. You needed to actually demonstrate your knowledge of the topic, and, in the case of specific problems like that one, you had to present solutions.
Thank you. The limit pops up a lot in engineering, not just on paper but in actual physical or electrical functions. That said, in electronics we use graph paper as writing paper so doing a graph with 3 samples is quicker.
This channel is way out of my league 99% of the concepts he deals with... but I still come back to watch more, lol. I like how he explains things and how he writes on his whiteboard.
I was taught that, if direct substitution results in A/B, where A and B are nonzero, that’s the limit. If you’re given a limit that is A/0, the limit DNE. If you’re given a limit that is 0/B, the limit is 0. If the limit is 0/0 it’s indeterminate. Methods like multiplying by conjugate, or L’Hopitals rule come in to play. So from first glance, you can instantly tell the limit DNE because the numerator is nonzero and the denominator is 0 when direct substitution is applied.
@@lugia8888 limit equal to positive or negative infinity is typically considered DNE though, right? Because it approaches different values from left and right.
@@wills4104 By definition, for a limit to exist in the first place, it must be a finite number. Both +∞ and -∞ aren't 'finite'. So when either the LHL or the RHL approaches either quantity, we say the limit doesn't exist.
@@wills4104not quite. a limit at ±inf can absolutely exist, just not in the space of real numbers. it is extremely common to extend the space of real numbers with ±inf when working with limits. in this case the limit still doesnt exist in that space because it approaches both +inf and -inf from either side, but you can just as well define an extension to the space with only one infinity. its also trivial to make a function that approaches +inf (or -inf) from both sides with a form like A/0
Btw, the graph shows that the two lines doesn't meet at a certain point (diverging on an asymptote) so that's what DNE literally means: there's no convergence at any particular point.
DNE actually means "does not exist" lol, a limit can still exist even if theres no convergence at any particular point, e.g. diverging to positive infinity/ negative infinity
@@bartiii7617 limits diverging to positive or negative infinity also do not exist by the delta epsilon definition, though most people just go with the "you know what i mean" equals sign
I just literally advance self-studying Calculus 1 rn and this is my 1st video yt recommend it. I didn't know that we can also have exponential signs to determine if + or - infinity but i alr knew that it will be DNE bcuz of + & - infinity are not equal. Perfect timing! I can't wait for my next sem. You got a sub❤!
This question gets a bit easier if you rewrite x/(x-1) as (x-1+1)/(x-1), which is 1 - 1/(x-1), which is a little more obvious in how it behaves as x -> 1. Fairly simple explanation though.
I just got started learning calculus, and I've been having a hard time understanding horizontal and vertical asymototpes, not sure if it's because the way my teacher teaches or some other reason. But I after clicking and watching this interesting video, you just made something click, thank you!
I still remember us being shocked when the teacher wrote positive and negative zero. We were perplexed, bamboozled even. Until he explained why and how.
Almost all limits can be evaluated by doing a thought experiment of "what happens if I move just ever so slightly to the left/right/both sides of it" - and then playing it out in your head.
brilliant explanation, these videos really make me feel like I'm getting a better grasp on calculus as someone who's never taken it but is passionate about math :)
When I took a course on it, lecturer said it doesn't exist (rather than its infinity), but on the video he's calling it infinity and -infinity, and for that reason the limit doesn't exist. I think I prefer saying it doesn't exist, but saying its infinity or -infinity gives you more insight into the shape of the graph I guess.
@@colinjava8447 The limit exists if and only if the right limit equals the left limit. If left limit is different from right limit (in a given point), the limit does not exist. The limit is not "either inf or -inf", it just "does not exist".
@@janskala22 I know, that's how I knew in 2 seconds that it doesn't exist (cause left =/= right). My point was in the video he writes infinity, when like you said it just doesn't exist. I think he knows that probably but does it for convenience.
@@colinjava8447 He only writes infinity on the right limit when it holds. He writes -infinity on the left limit where it holds. He does not write any definitive answer to the whole limit until he is sure it's DNE.
I personally prefer using two sequences to show that. X_n=1+-1/n, and then the functions turns to to 1+-n, and when you take its limit you get +-infinity. Feels more rigorous to me.
You can use graphical methods too: x = x - 1 + 1 ==> x/(x - 1) = 1 + 1/(x - 1). So y = x/(x - 1) is basically the 1/x graph shifted to right by 1 and up by 1 unit. As x -> 1, x/(x - 1) diverges. So limit does not exist if you know the 1/x graph well.
your teaching style is comforting i still dont understand this one due to my lack of foundational knowledge, i think. still very glad to have your vids
Please HELP me. - at 4:53 how did you say that 1- was still positive as 0.99999 ? I need answert or else I wont be able to sleep and I have no one who can explain me like this. Please help.
Well, every neighbourhood of 1 in the real line contains elements that are less than 1 but are greater than 0. Specifically, 0.99999 is less than 1, but is greater than 0.
From the times, when I was a student, I remember three different intinities: "+∞", "-∞" and "∞". So we explicitly used sign, if the infinity had one, and not used if that was "just the infinity", when sign is unknown (or does not matter). Is this the case nowadays? You are never using "+∞" notation, always omitting "+" sign...
The space that is assumed in the video is the standard extension of R with two points at infinity - +∞ and -∞. Unsigned ∞ does not exist in that space. I think it's a bad decision on the author's part to not explicitly state what space we are looking for a limit in, as in other extensions of R that limit does exist.
Yeah, this is a pretty basic video that doesn't enter into that, but that's exactly how we treat in class (I teach at University of Buenos Aires). Most time basic calculus students mess with the + or - infty and I tell to just rite infty symbol since we just search for vertical asymptotes only. So we just care on when the function collapses. Although there are some teachers who put enphasis on the sign, I don't agree to go into that when the students are struggling to more basic things at that level.
Nice video! It’s been a while since I’ve done this. Since we did indeed get the conclusion that we would expect from inspection, could you give an example of a limit that looks DNE at a glance, but turns out not to be?
@@literallyjustayoutubecomme1591 Agree. To beginning Calculus students, a limit often "looks DNE" as soon as they see that 0 in the denominator (even if the numerator also approaches 0).
As someone else stated, it sort of depends on how good your “glance.” Is. If your very proficient with limits and calculus, you potentially could have know just by looking at the limit what the answer would be. But a great example in this case would be x/(x-1)^2. Having a square term in the denominator actually causes the limit to approach positive infinity from both the left AND the right. Therefore the limit actually approach’s infinity and therefore does exist!
@@jackbrax7808 Depends on what you mean by "exist." According to most basic Calculus books I'm familiar with, if the limit is ∞, the limit doesn't exist-you're just being more specific about how/why it doesn't exist.
@@Steve_Stowers I just double checked my definitions and turns out your right. It doesn’t exist but both sides tend to infinity. But due to infinity not being a number, it doesn’t exist. But you can say the limit tends to infinity.
I did all this in my head. No joke. I know the number to the left of 1 minus 1 must be negative 0.00001 (something like that) and the number to the right of 1 minus 1 must be positive 0.00001 so I can include the right-handed limit approaches positive infinity and the left-handed limit approaches negative infinity, so the general limit does not exist. Easy peazy lemon squeezy.
We were taught that the general limit for that would not exist because one side goes to infinity and the other goes to negative infinity. You would have to do a directional limit
If the limit tends to infinity, it does not exist, because infinity is a concept, not a number. What infinity means in this context is that as we get closer to the limit, the value is always greater.
I personally think it's easier to approach these problems by changing the limiting value to 0. That way it's obvious what's +ve and -ve. In this case, we could change the limit to lim e->0 (1 + e) / e by substituting x for 1 + e (e is supposed to be epsilon here). Then you can work out lim e->+0 and e->-0 and it's a little easier (IMO).
Using precalculus and algebra 2 techniques we can easily determine the asymptote of the function and later the limit x/(x-1) we can use zero product property on (x-1) which will give us a vertical asymptote of 1 so when approaching x->1 u will get both infinity and -infinity showing that our limit DNE
If you have a function 'f' which is defined on a subset 'M' of real numbers and you have some real number 'y', then the left-side limit 'lim_{x->y-} f(x)' is defined as the limit 'lim_{x->y} g(x)' where 'g' is the same as 'f' but restricted to the subset of 'M' containing only the numbers that are at most 'y'. The right-side limit is defined analogous.
@@marvinliraDEThe point is, the limit of the function defined on ℝ\{1} doesn't exist. The problem is in the question (asking for something that doesn't exist,) not in the answer, for DNE is not the limit of the function.
@@clmasse I'm not sure there's anything wrong with asking a mathematical question for which there is no defined answer. Would you feel better if the question was "what is the limit of as x approaches , if such limit exists"? Because I'd think the qualification is implied for students beyond the most rudimentary level of maths.
but if we use L Hospital rule and differentiate the numerator and denominator then we have 1/1-0 which equals 1 so by that method the limit should exist and should be equal to one
'but if we use L Hospital rule' We get nothing, as L'Hopital (has nothing to do with hospitals) requires either of the indeterminate forms 0/0 and inf/inf. 'and differentiate the numerator and denominator then we have 1/1-0 which equals 1' And that is obviously incorrect, as even in the interval (1/2, 3/2) (which is a neighbourhood of 1) we have the infimum of |f(x)-1| = 2 for x in that interval and f(x) = x/(x-1), meaning that lim(f(x)) as x->1 cannot be 1.
While I do like the idea of the plus on the zero meaning a number arbitrarily close to zero, for problems like this I always think of it a “positive” zero. It’s functionally the same and gets the same answer, I just find it easier to understand, if you divide a positive by a positive, you get a positive. Is the zero positive or negative, not really, but if anyone is having trouble understanding this, try thinking of it this way.
It might help you with limits but it's not functionally the same. It's worth learning what 0+ actually means and sticking to that it will help you later with series etc
I dunno if its strictly valid, but my instinct is to transform it using fraction decomposition: x/(x-1)=1+1/(x-1) and substitution of t=x-1 to be: lim t→0 1+1/t Which is immediately obvious as undefined
I am absolutely certain that my Calculus teacher from 20ish years ago would have hated your 0+ notation…. She wanted derivative tests all the way. EDIT: To clarify, she would obviously have been fine with 0+ in the initial limit, but she wouldn't have liked 0+ as a result of partial computation. [Though it does seem intuitive as shown in this video.] She would have considered this an "invalid shortcut".
f' (x) := lim (h->0) (f(x+h) - f(x))/h well that's how I was taught how to find the derivative, using First Principles as it were. However when they teach Calc1 now they miss out this and expect you to look the derivative up in a table, usually supplied with the exam, whats the point of even learning calc this way! 😟
I always loved problems like this, it always reminded me that when u set up a number line the distance btw what ever numbers you end up choosing is infinite and if you wanted to count every number btw that distance you would always be approaching a certain number and never really reaching it.
i feel like the answer is either positive or negative infinity, but it's not defined from which side are we approaching x, limits tend to have positive and negative zeroes that gives 2 possible outcomes for 1 limit so it just doesn't work
in Russia the answer is actually infinity. Because we generally separate three infinities: -infinity, +infinity and infinity(point on infinite distance from zero like Complex Analysis style) in this case the answer is infinity, because for all epsilon>0 exists delta such that for x in [-delta,delta] the absolute value of x/(x-1) is greater that epsilon therefore the answer is INFINITY(not positive nor negative, just infinity)
That's fair. In complex analysis, the limit would equal the single unsigned ∞. But in real analysis, one side is +∞ and the other is -∞, and +∞≠-∞, so the limit fails to exist. And I assume you still acknowledge different limits that fail to exist. e^(1/x) on R. From the right, the limit is +∞. From the left, the limit is 0. +∞≠0 so the limit doesn't exist.
Man, my teacher has been hammering our class wuth this for months and bearly anyone understood even the priniples of "limit" and how you check for it. I'm just now trying to keep up with the material as we're quite past that and even had a small test (which I failed, miserably) and I'm going to definitely retake it soon as thanks to you I understand everything perfectly, even though English is not my first language haha Lots of love from Poland! Cheers!
'Man, my teacher has been hammering our class wuth this for months and bearly anyone understood even the priniples of "limit" and how you check for it' Let me guess, your teacher said that a limit is something that a function gets closer and closer to as its argument gets closer to some point? Yeah, I'd advice looking up an actual definition of a limit.
Back in Calculus 1 I got this wrong by multiplying by (x + 1)/(x + 1), giving me x(x + 1)/[(x-1)(x+1)] = (x^2 + x)/(x^2 - 1) Then I applied Le Hospital and got (2x + 1)/2x and came away with 3/2. Oops. *Also I will forever call L'Hôpital‘s rule Le Hospital.
Do you think a teacher or other person who check your work will be fine with 1+/0+ argumentation? I think better would be substitute y = (x - 1) so, we need to calc (y + 1) / y with y -> 0 then (y + 1)/y = 1 + 1/y, thus for y > 0 it (y + 1) / y > 1 / y, but 1/y -> infinity but for y < 0 it doesn't work. so I think just use some fact lim (x + C) = C + lim (x). idk.
It's how I was taught at uni. Best to check with your professor if you want to be sure. Don't forget you can always (and often should) add annotations in plain language explaining what you're doing and why. Then it doesn't really matter what notation you use, as long as it's clearly defined and consistently applied.
I really like this explanation since this explanation shows us that we mathematician did not do math recklessly according to the writing only, but according to the meaning of the limit hidden in the math problem.
If we just assume the standard extension of R with two points at infinity, then yes. If we don't, there is another fairly standard space where the limit does exist - the standard extension of R with one point at infinity.
-X approaches a vertical asymptope from the left, whereas +X approaches a vertical asymptope from the right. If -X doesn't equal +X the limit does not exist.
it is ±infinite depends on which side you take. - if approach from -inf to 1 + if approach inf to 1. You can easily check using scientific calculator. Type in the function and calculate wiith x= n±10^(-6)
'it is ±infinite depends on which side you take' lim(x/(x-1)) as x->1 considers points in the entire neighbourhood of 1. You are thinking of one-sided limits.
Nice, this is exactly how I did it. Can you post more proofs for common derivatives using the limit definition of (f(x+h) - f(x))/h? I think it could be fun to do a whole series on that. I tried the polynomial one for myself, and was able to confirm that (x^n)’, using binomial theorem and being left with just the second term was nx^(n-1). I tried getting the derivative of e^x = e^x, but I couldn’t pull it off tho, wasn’t sure how to bring out the h 😂
@@Syndicalism nice! I looked at the standard way I guess you could call it for evaluating the last part, where you say that some variable, eg k = the top part, so it becomes k->0 k/ln(k+1). Bringing the k up top to the bottom w reciprocal, and then log power rule it becomes 1/ln(e) = 1. The main part that was sort of unexpected for me was the start, setting the top to a variable. How might we stumble upon this-just trying it out bc it’s limit approaches 0 as well? Also do you think that mathematicians found out the derivatives first and then tasked themselves with proving them?
I approached it graphically. Let X divided by (X-1) equal Y, then start graphing a few points. For very large values of X, Y approaches 1 (from above). You're dividing large numbers by a number just slightly smaller. As X gets closer to 1 (from above), the graph turns up and heads toward positive infinity. Turn that around and for very negative values of X, you're dividing X by a number just slightly more negative than X. So Y approaches 1 (from below). As X gets closer to 1 (from below), Y gets smaller. It crosses the origin at X=0, then goes negative and heads off to negative infinity. Been too long for me to recall if this is a hyperbola or something else. But it's still a fun little question, and an important light bulb to turn on in students.
@@isaacbruner65 To say that the limit does not exist is just another way of saying that there is no solution to the problem of finding the limit. The statement that a solution does not exist is not in itself a solution. If it were, then you could say that every equation has a solution, which makes a nonsense of the concept of a solution.
@@omp199it's different with limits. "The limit does not exist" is part of the set of possible solution with limits. Similar to NaN is a possible answer to a floating point operation even though NaN literally means Not a Number.
@@kazedcat No. It's not "different with limits". A solution is a value or set of values that satisfy a given set of conditions. If the condition is that of being the limit of an expression, then the nonexistence of a limit implies the nonexistence of a solution. As for "Nan", you are bringing programming language conventions into a discussion of mathematics. A programming language might have a function that returns NaN in certain circumstances, but that has nothing to do with mathematics.
my first thought it should be the same as lim (x->0)1/x (substract 1 and add 1in the numerator, simplify to lim(x->1) 1 + 1/(x-1), 1 is a constant and don't really change anything here and finally lim(x->1) 1/(x-1) looks almost like lim (x->0) 1/x, which explanation, usually always given in a class rooms).
@@isaacbruner65 No, it would not. There are multiple extensions of R, one of which is assumed by BPRP in the video and has two points at infinity, and another one has only one point at infinity, which we can call 'unsigned infinity' for clarity's sake. In the case of the latter one, the person you responded to is absolutely correct, and it is a bad thing that BPRP did not explicitly bring up the matter of the space in which we are supposed to look for a limit.
Also, even if both the 1+ and 1- limits became the same(∞ or -∞), the limit would still not exist because limit only exists when Left hand limit= right hand limit= finite quantity
Damn I wasn’t taught that there are two answers. I just said positive infinity instantly without knowing that you have two answers , both negative and positive, and have to check left and right side to see which one is correct
You can also answer lim_x→1[x/(x-1)]=∞ without precising the sign, and it means that there is an asymptote at x=1 in this case, without telling from where the function approaches it. Be careful not to mistake it for lim=+∞, tho, this one would mean that the limit is defined. This is also why I would recommend to always say +∞ instead of just ∞ when the value is positive.
Like square roots, common practice has people being sloppy. They don't do much to distinguish, for example, positive infinity from unsigned infinity, or the multivalued square root from a single branch (generally the principal one).
The left right method is unnecessary to me, if the expression is unbounded as x->a then you're done. The left right method won't work on limits like lim_{x->0} 1/x^2 anyway.
The hardest thing for me was, and still is, to understand how we can claim that the limit does not exist, when we have just proved that it equals to [-∞, +∞], so it does exist. Be it in such an exotic form. Would anyone mind explaining?
U can also take For right hand limit x=1+h where h-->>0 And for left hand limit x=1-h where h-->>0 For limit to exist, Right hand limit = Left Hand limit In this question RHL=+2, LHL=-2 therefore limit doesn't exist
'U can also take For right hand limit x=1+h where h-->>0 And for left hand limit x=1-h where h-->>0 ' No, you can't. You have to use h->0+ and h->0-. Otherwise, you are just looking at a limit as x->1 either way. 'For limit to exist, Right hand limit = Left Hand limit' Strictly speaking, that's not true. That depends on the topology of the space where we are looking for a limit in. 'In this question RHL=+2, LHL=-2 therefore limit doesn't exist' In the case in the video, with lim(x/(x-1)) as x->1, assuming that we are looking for a limit in the standard two-point compactification of the real line, the RHL is +infinity, and LHL is -infinity, as shown in the video.
Happy not to find an "obvious solution" like Infinity or so. After 11 years of my first calculus class, I'm careful enough not to find anything trivial hahaha
'limit of (1+ε)/ε as ε→0' Does not exist in the two-point compactification of the real line. 'which is ±∞ depending whether ε0' The expression 'as ε→0' means that we go over the entire punctured neighbourhoods U(0)\{0} , and not just their portions that are greater or less than 0.
Good old math yay! I simply said "negative infinity" since I only considered x approaching 1 as x = 0.99999999, but yeah it could have been x = 1.000000001 too.
My confusion is when you evaluate 1+. My understanding of “limit as X -> 1,” is that X gets very very close, but not quite 1. I’m confused why X would go past the limit.
This is incredible. I cannot believe it. This has got to be some kind of a record. A miracle. I tried it and I got it right. I never get things right. This got to be the exception to prove the rule. Wait...
I once drew a piecewise function for my learners, below 2 it was defined as x and from 2 and above it was defined as x+2, I asked them to determine the limit of the function as it approaches 2. I realized they had long forgotten about approaching a point from the left and right, they've gotten used to simplifying then "substituting" the limit value.
This problem can only be solved by making the 1 a 1+ or 1-, giving us a way to represent limits going toward infinity. Without the qualifier of + or - to show which infinity we are going toward, the problem doen't even exist. 1/1-0 is literally going nowhere because there are no directions to take. 1/1-0 is the mathematician's Schrödinger's Cat.
'This problem can only be solved by making the 1 a 1+ or 1-' 'Without the qualifier of + or - to show which infinity we are going toward, the problem doen't even exist' No, this problem is perfectly solvable. Just watch the video. Also, the expressions 'x->x_0+' and 'x->x_0-' do not refer to the tendency of something else towards any particular points at infinity. Those expressions simply tell us what portions of neighbourhoods of x_0 we are considering.
f(x) = x/(x- 1) = 1+ 1/(x-1) let's consider 1/x and 1/(x-1) A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift 1/(x-1) is a horizontal shift of the function 1/x, the graph will shift right The simplest shift is a vertical shift, moving the graph up because this transformation involves adding a positive constant to the function 1/(x-1).
Why doesn’t this limit represent the definition of derivative? 👉ua-cam.com/video/C440uWSzFGg/v-deo.htmlsi=Zx-wuAEkbfV5_m8a
This immediately reminded me to when I was thought the limit of 1/x as x goes to 0 to illustrate why one can't divide by 0.
but if we use L Hospital rule and differentiate the numerator and denominator then we have 1/1-0 which equals 1 so by that method the limit should exist and should be equal to one
@@winners-r4z Not familiar with the L Hospital rule, So I wonder how you got from x/(x-1) to 1/(1-0) or 1/1-0.
@@winners-r4z We can only apply the L'Hospital's rule if the direct substitution returns an indeterminate form, that means 0/0 or ±∞/±∞.
As bprp said, this is *NOT* an undeterminate form. The limit is clearly ∞, but depending on the direction from which we approach 1, the sign of ∞ changes
Brilliant. I like this kind of videos. Subscribed🎉
Honestly I think the class didn’t learn limit from right and left or they just forget about it.
Well, in both cases the function approaches 1/0. However, where x is less than 1 and greater than 0, the function is negative. Likewise, when x is greater than 1 or less than 0 (which is irrelevant to this question), the function is positive.
I did learn about that (albeit 25 years ago, oh god, and not with this guy’s notation) and my first reaction was that it doesn’t specify from which side it is in the problem.
@@JasperJanssen
Why is that a problem? If the side is not specified, it's obviously the standard, and not a one-sided limit.
@@thetaomegatheta … did you watch the video? And no, there is no such thing as “the standard”.
@@JasperJanssen
' … did you watch the video?'
Yes, I did.
'And no, there is no such thing as “the standard”
Do you seriously not know about the non-one-sided limits?
This limit problem is a good illustration of why making even a rough sketch graph of the function in question can shed a lot of light. Using a graphing as a qualitative analytical tool is too often overlooked.
While i do agree that graphs are amazingly helpful,i believe more complex problems would be better suited for them,idk.. to me, the answer felt glaringly obvious from the start.
@@levaniandgiorgi2358 I largely agree. I looked at the statement and pretty much saw the answer immediately, but then I have the advantage over freshman students of having done math for close to 60 years. I’ve encouraged students to not shy away from sketching Bode plots or pole/zero diagrams in the EE courses I’ve taught. It is handy to look at a problem with more than one method to avoid mistakes. The backup method doesn’t need to be precise, just accurate enough to confirm your thinking is on track.
No reason to waste time graphing something like this
Graphs do what Graphs are supposed to do, give you a visual representation of the abstract equation
It's helpful for people who are stronger visual learners to link the reasoning and the answer together
Graphs are very powerful. It's really hard to believe calculus was invented without using them. People make fun of 'trivial' stuff like rolle's theorem, but good luck proving them without graphs.
Note to self:
"L'Hôpital's rule can only be applied in the case where direct substitution yields an indeterminate form, meaning 0/0 or ±∞/±∞. So if f and g are defined, L'Hôpital would be applicable only if the value of both f and g is 0. "
I made the same mistake
same mistake here😢
Same and got answer one
If it's a school or university question, L'Hôpital's rule shouldn't be accepted as a valid proof to be honest
@@vintovkasnipera Why's that?
I used to love asking my students on these if their denominator was positive zero or negative zero. The transition from initial confusion to a-ha was one of my favorite gems from teaching.
Sorry, I don't understand what positive and negative zero mean. Could you please explain?
@@krishnannarayanan8819shorthand for "approaches zero from the positive direction" and negative directing, respectively
@@krishnannarayanan8819 A shorthand way of saying to approach 0 from x < 0, or x > 0.
@@krishnannarayanan8819it means 0 or -0
@@krishnannarayanan8819 you could assume some number "h" which is a very small positive number. positive 0 means 0+h and negative 0 means 0-h . basically 0+ and 0- are approaching 0 from right and left sides respectively.
Cady Heron would have figured this out… it’s how she won the athlete competition in “Mean Girls.”
I love the movie reference.
I just woke up and this video was suggested. I haven't touched in limits for almost a decade so my thought was "Ive forgetten all of this".
I've opened the video,watched for 3 minutes and I could feel the knowledge coming back 😂 so weird
Thats how most people say videos are teaching more than school
In reality its just bringing stuff back.
😂😂😂😂😂🤗🤗🤗
it feels like an old rusty gear inside your head has started to turn
I'm still worthy feeling 😂😂
So I guess that your knowledge has limits.
never took a calculus class in my life but i still end up watching these videos
took multiple calc DQ and didn't really like it, but watch
studying in class 10 but still watched cuz why not 😂
Admirable curiousity. 👍
@@jamescollier3 I never really liked math as a whole in college. Then once I graduated, I started learning it in depth, on my own . Then I started loving it.
Now it has been 7 years since I graduated and I still learn it
My calc class explained SOOOO MANY of the questions I had about things that didn't make sense from earlier classes. Can recommend.
Are we not gonna talk about the amount of markers he's got stored in the background? I've seen entire schools have less than that lmao
He just buys the whole stock 😭😭
Knowing how this function's graph behaves gives all of the intuition needed. Vertical asymptote at x=1, positive to the right, negative between 0 and 1. My students often dive into the calculus without thinking about the precalculus. Sure it can be done without the precalc, but the confidence gets a big boost when we think about the graph first.
And that's how you get 0 points for the question on an exam. You're expected to do the calculus on a calculus exam.
@@No-cg9kj read more carefully
Approaching a limit doesn't require a Y-axis, you're needlessly complicating the concept and conditioning them with a euclidian bias in their thinking about numbers.
@No-cg9kj On any calculus exam, this question would be a multiple-choice or short answer question where you’re either right or wrong, no partial credit. Graphing is sometimes way faster than doing the math. just visualizing the graph i solved this problem in probably 2-3 seconds
@@iamcoolkinda
'On any calculus exam, this question would be a multiple-choice or short answer question where you’re either right or wrong, no partial credit'
Literally none of the math exams that I took at university had multiple-choice questions. You needed to actually demonstrate your knowledge of the topic, and, in the case of specific problems like that one, you had to present solutions.
I haven't been to school for at least 5 years, and now listening to your explanation I could understand better what I could not back then
Thank you. The limit pops up a lot in engineering, not just on paper but in actual physical or electrical functions.
That said, in electronics we use graph paper as writing paper so doing a graph with 3 samples is quicker.
Finally! A problem on this channel I could solve on my own!
This channel is way out of my league 99% of the concepts he deals with... but I still come back to watch more, lol. I like how he explains things and how he writes on his whiteboard.
I love this bite sized math content, feels like I'm getting a bit smarter every day :D
I was taught that, if direct substitution results in A/B, where A and B are nonzero, that’s the limit.
If you’re given a limit that is A/0, the limit DNE.
If you’re given a limit that is 0/B, the limit is 0.
If the limit is 0/0 it’s indeterminate. Methods like multiplying by conjugate, or L’Hopitals rule come in to play.
So from first glance, you can instantly tell the limit DNE because the numerator is nonzero and the denominator is 0 when direct substitution is applied.
You can have a limit equal to positive or negative infinity. Also, aside from Hopital you can use Taylor series.
@@lugia8888 limit equal to positive or negative infinity is typically considered DNE though, right? Because it approaches different values from left and right.
@@wills4104
By definition, for a limit to exist in the first place, it must be a finite number. Both +∞ and -∞ aren't 'finite'. So when either the LHL or the RHL approaches either quantity, we say the limit doesn't exist.
@@wills4104not quite. a limit at ±inf can absolutely exist, just not in the space of real numbers. it is extremely common to extend the space of real numbers with ±inf when working with limits. in this case the limit still doesnt exist in that space because it approaches both +inf and -inf from either side, but you can just as well define an extension to the space with only one infinity. its also trivial to make a function that approaches +inf (or -inf) from both sides with a form like A/0
Btw, the graph shows that the two lines doesn't meet at a certain point (diverging on an asymptote) so that's what DNE literally means: there's no convergence at any particular point.
DNE actually means "does not exist" lol, a limit can still exist even if theres no convergence at any particular point, e.g. diverging to positive infinity/ negative infinity
maybe its TNCAAPP: "theres no covergence at any point"
@@bartiii7617 limits diverging to positive or negative infinity also do not exist by the delta epsilon definition, though most people just go with the "you know what i mean" equals sign
@@hyperpsych6483can't you use the epsilon-N/delta-M/N-M definitions for those limits?
@@hyperpsych6483I think it depends on the topology we’re working on, tho in common topology of ℝ we regard ±∞ as DNE, so I agree with you
I just literally advance self-studying Calculus 1 rn and this is my 1st video yt recommend it. I didn't know that we can also have exponential signs to determine if + or - infinity but i alr knew that it will be DNE bcuz of + & - infinity are not equal. Perfect timing! I can't wait for my next sem. You got a sub❤!
Glad to hear! Thank you!
Im grade 7(ph) and i understand calculus :)))))
Is this a well-defined limit? Calculus question on Reddit r/askmath
ua-cam.com/video/WaFEqnPGzCo/v-deo.html
This question gets a bit easier if you rewrite x/(x-1) as (x-1+1)/(x-1), which is 1 - 1/(x-1), which is a little more obvious in how it behaves as x -> 1. Fairly simple explanation though.
Meh
Is there no fancy way like multiplying with x+1 or the usual trickery for limits I never really learned well enough to understand?
No. Try it yourself and see.
I just got started learning calculus, and I've been having a hard time understanding horizontal and vertical asymototpes, not sure if it's because the way my teacher teaches or some other reason. But I after clicking and watching this interesting video, you just made something click, thank you!
I haven't taken calculus in my life and was interested. Only to be completely distracted by the lifetime supply of expo markers.
I still remember us being shocked when the teacher wrote positive and negative zero.
We were perplexed, bamboozled even. Until he explained why and how.
Every time i see a limit i just l'hosptal the thang and pray for the best 😭😭😭
Get your indeterminate cat t-shirt: 👉 amzn.to/3qBeuw6
Almost all limits can be evaluated by doing a thought experiment of "what happens if I move just ever so slightly to the left/right/both sides of it" - and then playing it out in your head.
This brings me back 50 years. Thanks.
I recently talked with one of the guys who was also in that class.
I was not getting LHL = RHL so I knew the limit does not exist. Btw Merry Christmas bprp :D
brilliant explanation, these videos really make me feel like I'm getting a better grasp on calculus as someone who's never taken it but is passionate about math :)
Merry Christmas Bprp 🎉
Thank you! You too!
I've found another method. We know that x/(x - 1) = ( (x - 1) + 1 )/(x - 1) = 1 + 1/(x - 1). So the limit equals 1 + lim_(x -> 1) { 1/(x - 1) }, or just 1 + lim_(x -> 0) { 1/x } which we know DNE.
When I took a course on it, lecturer said it doesn't exist (rather than its infinity), but on the video he's calling it infinity and -infinity, and for that reason the limit doesn't exist.
I think I prefer saying it doesn't exist, but saying its infinity or -infinity gives you more insight into the shape of the graph I guess.
@@colinjava8447 The limit exists if and only if the right limit equals the left limit. If left limit is different from right limit (in a given point), the limit does not exist. The limit is not "either inf or -inf", it just "does not exist".
@@janskala22 I know, that's how I knew in 2 seconds that it doesn't exist (cause left =/= right).
My point was in the video he writes infinity, when like you said it just doesn't exist.
I think he knows that probably but does it for convenience.
@@colinjava8447 He only writes infinity on the right limit when it holds. He writes -infinity on the left limit where it holds. He does not write any definitive answer to the whole limit until he is sure it's DNE.
@@janskala22 I know, I saw the video too.
I personally prefer using two sequences to show that. X_n=1+-1/n, and then the functions turns to to 1+-n, and when you take its limit you get +-infinity.
Feels more rigorous to me.
You can use graphical methods too: x = x - 1 + 1 ==> x/(x - 1) = 1 + 1/(x - 1). So y = x/(x - 1) is basically the 1/x graph shifted to right by 1 and up by 1 unit. As x -> 1, x/(x - 1) diverges. So limit does not exist if you know the 1/x graph well.
your teaching style is comforting
i still dont understand this one due to my lack of foundational knowledge, i think.
still very glad to have your vids
It's been a while since I did this type of stuff. Thanks for the refresher.
Please HELP me. - at 4:53 how did you say that 1- was still positive as 0.99999 ? I need answert or else I wont be able to sleep and I have no one who can explain me like this. Please help.
Well, every neighbourhood of 1 in the real line contains elements that are less than 1 but are greater than 0.
Specifically, 0.99999 is less than 1, but is greater than 0.
Imagine a number very slightly lower than 1. Such as 0.9999. That's still greater than 0.
1- is the number before 1
Or just x/(x-1) = 1+1/(x-1) so it's 1+ (lim x-> 0 1/x) which is dne
From the times, when I was a student, I remember three different intinities: "+∞", "-∞" and "∞". So we explicitly used sign, if the infinity had one, and not used if that was "just the infinity", when sign is unknown (or does not matter). Is this the case nowadays? You are never using "+∞" notation, always omitting "+" sign...
Same
The space that is assumed in the video is the standard extension of R with two points at infinity - +∞ and -∞. Unsigned ∞ does not exist in that space.
I think it's a bad decision on the author's part to not explicitly state what space we are looking for a limit in, as in other extensions of R that limit does exist.
Yeah, this is a pretty basic video that doesn't enter into that, but that's exactly how we treat in class (I teach at University of Buenos Aires). Most time basic calculus students mess with the + or - infty and I tell to just rite infty symbol since we just search for vertical asymptotes only. So we just care on when the function collapses. Although there are some teachers who put enphasis on the sign, I don't agree to go into that when the students are struggling to more basic things at that level.
I immediately thought about which direction when I saw x->1 and x-1
Nice video! It’s been a while since I’ve done this. Since we did indeed get the conclusion that we would expect from inspection, could you give an example of a limit that looks DNE at a glance, but turns out not to be?
Well, that depends on how good your glances are, doesn’t it :)
@@literallyjustayoutubecomme1591 Agree. To beginning Calculus students, a limit often "looks DNE" as soon as they see that 0 in the denominator (even if the numerator also approaches 0).
As someone else stated, it sort of depends on how good your “glance.” Is. If your very proficient with limits and calculus, you potentially could have know just by looking at the limit what the answer would be.
But a great example in this case would be x/(x-1)^2. Having a square term in the denominator actually causes the limit to approach positive infinity from both the left AND the right. Therefore the limit actually approach’s infinity and therefore does exist!
@@jackbrax7808 Depends on what you mean by "exist." According to most basic Calculus books I'm familiar with, if the limit is ∞, the limit doesn't exist-you're just being more specific about how/why it doesn't exist.
@@Steve_Stowers I just double checked my definitions and turns out your right. It doesn’t exist but both sides tend to infinity. But due to infinity not being a number, it doesn’t exist. But you can say the limit tends to infinity.
I did all this in my head. No joke. I know the number to the left of 1 minus 1 must be negative 0.00001 (something like that) and the number to the right of 1 minus 1 must be positive 0.00001 so I can include the right-handed limit approaches positive infinity and the left-handed limit approaches negative infinity, so the general limit does not exist. Easy peazy lemon squeezy.
We were taught that the general limit for that would not exist because one side goes to infinity and the other goes to negative infinity. You would have to do a directional limit
Everybody gangster until g(x)=0
-_-
If the limit tends to infinity, it does not exist, because infinity is a concept, not a number. What infinity means in this context is that as we get closer to the limit, the value is always greater.
I personally think it's easier to approach these problems by changing the limiting value to 0. That way it's obvious what's +ve and -ve. In this case, we could change the limit to lim e->0 (1 + e) / e by substituting x for 1 + e (e is supposed to be epsilon here). Then you can work out lim e->+0 and e->-0 and it's a little easier (IMO).
Using precalculus and algebra 2 techniques we can easily determine the asymptote of the function and later the limit
x/(x-1)
we can use zero product property on (x-1) which will give us a vertical asymptote of 1 so when approaching x->1 u will get both infinity and -infinity showing that our limit DNE
It's normal, there is no limit. Ask the good question: what is the + or the - limit? The problem is often the nut behind the whiteboard.
If you have a function 'f' which is defined on a subset 'M' of real numbers and you have some real number 'y', then the left-side limit 'lim_{x->y-} f(x)' is defined as the limit 'lim_{x->y} g(x)' where 'g' is the same as 'f' but restricted to the subset of 'M' containing only the numbers that are at most 'y'. The right-side limit is defined analogous.
So visually speaking, you cut your function at the point 'y' into a left side and a right side and handle each side on its own.
how did u know i nut behind the whiteboard? also he answered those as + and - infinity already.
@@marvinliraDEThe point is, the limit of the function defined on ℝ\{1} doesn't exist. The problem is in the question (asking for something that doesn't exist,) not in the answer, for DNE is not the limit of the function.
@@clmasse I'm not sure there's anything wrong with asking a mathematical question for which there is no defined answer.
Would you feel better if the question was "what is the limit of as x approaches , if such limit exists"? Because I'd think the qualification is implied for students beyond the most rudimentary level of maths.
but if we use L Hospital rule and differentiate the numerator and denominator then we have 1/1-0 which equals 1 so by that method the limit should exist and should be equal to one
'but if we use L Hospital rule'
We get nothing, as L'Hopital (has nothing to do with hospitals) requires either of the indeterminate forms 0/0 and inf/inf.
'and differentiate the numerator and denominator then we have 1/1-0 which equals 1'
And that is obviously incorrect, as even in the interval (1/2, 3/2) (which is a neighbourhood of 1) we have the infimum of |f(x)-1| = 2 for x in that interval and f(x) = x/(x-1), meaning that lim(f(x)) as x->1 cannot be 1.
Dangg, thanks a lot dude@@thetaomegatheta
just after reading that first line of your comment i was like " ahh shit , i forgot bout that"
XD
I guessed that it was undefined in 2 seconds, cause its essentially 1/x, and its a limit from both sides.
Why? substitute s=x-1 immediately yields x/(x-1) = s/s+1/s as s->0+ positive inf, s->0- negative inf. Therefore undefined
While I do like the idea of the plus on the zero meaning a number arbitrarily close to zero, for problems like this I always think of it a “positive” zero. It’s functionally the same and gets the same answer, I just find it easier to understand, if you divide a positive by a positive, you get a positive. Is the zero positive or negative, not really, but if anyone is having trouble understanding this, try thinking of it this way.
Huh, that actually is a really nice way of thinking of it. Thanks for this!
It might help you with limits but it's not functionally the same. It's worth learning what 0+ actually means and sticking to that it will help you later with series etc
The right usage of infinity is a gulp of fresh air.
5:38 look at the asymptote on that mother function... Don't you have that t-shirt?
what is "mother function" ?
I dunno if its strictly valid, but my instinct is to transform it using fraction decomposition:
x/(x-1)=1+1/(x-1)
and substitution of t=x-1 to be:
lim t→0 1+1/t
Which is immediately obvious as undefined
I am absolutely certain that my Calculus teacher from 20ish years ago would have hated your 0+ notation…. She wanted derivative tests all the way.
EDIT: To clarify, she would obviously have been fine with 0+ in the initial limit, but she wouldn't have liked 0+ as a result of partial computation. [Though it does seem intuitive as shown in this video.] She would have considered this an "invalid shortcut".
f' (x) := lim (h->0) (f(x+h) - f(x))/h
well that's how I was taught how to find the derivative, using First Principles as it were.
However when they teach Calc1 now they miss out this and expect you to look the derivative up in a table, usually supplied with the exam, whats the point of even learning calc this way! 😟
1/0 is not allowed.
1+ and 1- are not numbers.
1/x has a singularity at x=0.
This singularity is shifted in his example.
If you've done the x/x limit before, it's super easy--graph looks the same just shifted 1 unit to the right due to the "-1" in the denominator.
I always loved problems like this, it always reminded me that when u set up a number line the distance btw what ever numbers you end up choosing is infinite and if you wanted to count every number btw that distance you would always be approaching a certain number and never really reaching it.
i feel like the answer is either positive or negative infinity, but it's not defined from which side are we approaching x, limits tend to have positive and negative zeroes
that gives 2 possible outcomes for 1 limit so it just doesn't work
Good video as always
in Russia the answer is actually infinity. Because we generally separate three infinities: -infinity, +infinity and infinity(point on infinite distance from zero like Complex Analysis style)
in this case the answer is infinity, because for all epsilon>0 exists delta such that for x in [-delta,delta] the absolute value of x/(x-1) is greater that epsilon therefore the answer is INFINITY(not positive nor negative, just infinity)
That's fair. In complex analysis, the limit would equal the single unsigned ∞. But in real analysis, one side is +∞ and the other is -∞, and +∞≠-∞, so the limit fails to exist.
And I assume you still acknowledge different limits that fail to exist. e^(1/x) on R. From the right, the limit is +∞. From the left, the limit is 0. +∞≠0 so the limit doesn't exist.
Man, my teacher has been hammering our class wuth this for months and bearly anyone understood even the priniples of "limit" and how you check for it.
I'm just now trying to keep up with the material as we're quite past that and even had a small test (which I failed, miserably) and I'm going to definitely retake it soon as thanks to you I understand everything perfectly, even though English is not my first language haha
Lots of love from Poland! Cheers!
'Man, my teacher has been hammering our class wuth this for months and bearly anyone understood even the priniples of "limit" and how you check for it'
Let me guess, your teacher said that a limit is something that a function gets closer and closer to as its argument gets closer to some point?
Yeah, I'd advice looking up an actual definition of a limit.
Back in Calculus 1 I got this wrong by multiplying by (x + 1)/(x + 1), giving me x(x + 1)/[(x-1)(x+1)] = (x^2 + x)/(x^2 - 1)
Then I applied Le Hospital and got (2x + 1)/2x and came away with 3/2. Oops.
*Also I will forever call L'Hôpital‘s rule Le Hospital.
Of course I later realized 1/0, as mentioned in the video, didn’t satisfy the criteria necessary to use Le Hospital’s rule.
I call it eel( yes the animal) hospital
Do you think a teacher or other person who check your work will be fine with 1+/0+ argumentation?
I think better would be substitute y = (x - 1) so, we need to calc (y + 1) / y with y -> 0
then (y + 1)/y = 1 + 1/y, thus for y > 0 it (y + 1) / y > 1 / y, but 1/y -> infinity
but for y < 0 it doesn't work. so I think just use some fact lim (x + C) = C + lim (x). idk.
It's how I was taught at uni. Best to check with your professor if you want to be sure.
Don't forget you can always (and often should) add annotations in plain language explaining what you're doing and why. Then it doesn't really matter what notation you use, as long as it's clearly defined and consistently applied.
@@dielaughing73 our professor also uses these notations what the hell is wrong with it
This is how I was taught as well. If a teacher or other person who will check your work isn't fine with it, it's time to talk to their boss.
Just replace 1+ with 1+ε and 1- with 1-ε.
@@billmilligan7272thanks for your input Karen
I really like this explanation since this explanation shows us that we mathematician did not do math recklessly according to the writing only, but according to the meaning of the limit hidden in the math problem.
DNE since f has a VA at x = 1 and f is odd
Vertical asymptote?
If we just assume the standard extension of R with two points at infinity, then yes.
If we don't, there is another fairly standard space where the limit does exist - the standard extension of R with one point at infinity.
f is not odd at all.
-X approaches a vertical asymptope from the left, whereas +X approaches a vertical asymptope from the right. If -X doesn't equal +X the limit does not exist.
Why L hopital rule is not used here
We don’t have an indeterminate form.
Because L’hôpital rule only works with 0/0 or infinity/infinity
Understood
Because the injury isn't bad enough to go to L'Hospital.
@@teelo12000 This isn’t ‘la Páris’! 😂😂😂😂😂😂😂😂😂😂😂😂
it is ±infinite depends on which side you take.
- if approach from -inf to 1
+ if approach inf to 1.
You can easily check using scientific calculator. Type in the function and calculate wiith x= n±10^(-6)
'it is ±infinite depends on which side you take'
lim(x/(x-1)) as x->1 considers points in the entire neighbourhood of 1. You are thinking of one-sided limits.
Nice, this is exactly how I did it. Can you post more proofs for common derivatives using the limit definition of (f(x+h) - f(x))/h? I think it could be fun to do a whole series on that. I tried the polynomial one for myself, and was able to confirm that (x^n)’, using binomial theorem and being left with just the second term was nx^(n-1). I tried getting the derivative of e^x = e^x, but I couldn’t pull it off tho, wasn’t sure how to bring out the h 😂
exp(x) is factored out of the limit. The remaining limit is [exp(h)-1]/h which evaluates to 1.
@@Syndicalism nice! I looked at the standard way I guess you could call it for evaluating the last part, where you say that some variable, eg k = the top part, so it becomes k->0 k/ln(k+1). Bringing the k up top to the bottom w reciprocal, and then log power rule it becomes 1/ln(e) = 1.
The main part that was sort of unexpected for me was the start, setting the top to a variable. How might we stumble upon this-just trying it out bc it’s limit approaches 0 as well? Also do you think that mathematicians found out the derivatives first and then tasked themselves with proving them?
I approached it graphically. Let X divided by (X-1) equal Y, then start graphing a few points. For very large values of X, Y approaches 1 (from above). You're dividing large numbers by a number just slightly smaller. As X gets closer to 1 (from above), the graph turns up and heads toward positive infinity. Turn that around and for very negative values of X, you're dividing X by a number just slightly more negative than X. So Y approaches 1 (from below). As X gets closer to 1 (from below), Y gets smaller. It crosses the origin at X=0, then goes negative and heads off to negative infinity.
Been too long for me to recall if this is a hyperbola or something else. But it's still a fun little question, and an important light bulb to turn on in students.
Of course class couldn't figure it out if no solution exists 😂
There is a solution and the solution is that the limit does not exist.
@@isaacbruner65 To say that the limit does not exist is just another way of saying that there is no solution to the problem of finding the limit. The statement that a solution does not exist is not in itself a solution. If it were, then you could say that every equation has a solution, which makes a nonsense of the concept of a solution.
@@omp199it's different with limits. "The limit does not exist" is part of the set of possible solution with limits. Similar to NaN is a possible answer to a floating point operation even though NaN literally means Not a Number.
@@kazedcat No. It's not "different with limits". A solution is a value or set of values that satisfy a given set of conditions. If the condition is that of being the limit of an expression, then the nonexistence of a limit implies the nonexistence of a solution.
As for "Nan", you are bringing programming language conventions into a discussion of mathematics. A programming language might have a function that returns NaN in certain circumstances, but that has nothing to do with mathematics.
@@omp199 Programming is mathematics. The Turing Machine is a mathematical object.
my first thought it should be the same as lim (x->0)1/x (substract 1 and add 1in the numerator, simplify to lim(x->1) 1 + 1/(x-1), 1 is a constant and don't really change anything here and finally lim(x->1) 1/(x-1) looks almost like lim (x->0) 1/x, which explanation, usually always given in a class rooms).
Looks like it just tends to ∞
That would imply that it tends to positive infinity which is obviously not the case.
@@isaacbruner65 🤓
@@isaacbruner65
No, it would not.
There are multiple extensions of R, one of which is assumed by BPRP in the video and has two points at infinity, and another one has only one point at infinity, which we can call 'unsigned infinity' for clarity's sake. In the case of the latter one, the person you responded to is absolutely correct, and it is a bad thing that BPRP did not explicitly bring up the matter of the space in which we are supposed to look for a limit.
@@lucaspanto9650 Your dumbass said the limit tends to infinity. I doubt you're in a position to use that emoji.
Negative infinity makes sense to me only needed to put in a couple of values to work it out. Been over 10 years since I've touched limits
It’s a two sided limit
I figured out it DNE in 5s lol
Do you want a sticker?
@@General12th lol
Also, even if both the 1+ and 1- limits became the same(∞ or -∞), the limit would still not exist because limit only exists when Left hand limit= right hand limit= finite quantity
Damn I wasn’t taught that there are two answers. I just said positive infinity instantly without knowing that you have two answers , both negative and positive, and have to check left and right side to see which one is correct
We always learned to take the limit approaching form the left, the right and then combine them
It’s only weird if you make x equal to 1, but we’re actually getting as close as possible to 1, but not 1.
You can also answer lim_x→1[x/(x-1)]=∞ without precising the sign, and it means that there is an asymptote at x=1 in this case, without telling from where the function approaches it. Be careful not to mistake it for lim=+∞, tho, this one would mean that the limit is defined.
This is also why I would recommend to always say +∞ instead of just ∞ when the value is positive.
Like square roots, common practice has people being sloppy. They don't do much to distinguish, for example, positive infinity from unsigned infinity, or the multivalued square root from a single branch (generally the principal one).
The left right method is unnecessary to me, if the expression is unbounded as x->a then you're done. The left right method won't work on limits like lim_{x->0} 1/x^2 anyway.
The hardest thing for me was, and still is, to understand how we can claim that the limit does not exist, when we have just proved that it equals to [-∞, +∞], so it does exist. Be it in such an exotic form.
Would anyone mind explaining?
U can also take
For right hand limit x=1+h where h-->>0
And for left hand limit x=1-h where h-->>0
For limit to exist, Right hand limit = Left Hand limit
In this question RHL=+2, LHL=-2 therefore limit doesn't exist
'U can also take
For right hand limit x=1+h where h-->>0
And for left hand limit x=1-h where h-->>0 '
No, you can't. You have to use h->0+ and h->0-. Otherwise, you are just looking at a limit as x->1 either way.
'For limit to exist, Right hand limit = Left Hand limit'
Strictly speaking, that's not true. That depends on the topology of the space where we are looking for a limit in.
'In this question RHL=+2, LHL=-2 therefore limit doesn't exist'
In the case in the video, with lim(x/(x-1)) as x->1, assuming that we are looking for a limit in the standard two-point compactification of the real line, the RHL is +infinity, and LHL is -infinity, as shown in the video.
Happy not to find an "obvious solution" like Infinity or so. After 11 years of my first calculus class, I'm careful enough not to find anything trivial hahaha
Let x=1+ε and take the limit of (1+ε)/ε as ε→0, which is ±∞ depending whether ε0.
'limit of (1+ε)/ε as ε→0'
Does not exist in the two-point compactification of the real line.
'which is ±∞ depending whether ε0'
The expression 'as ε→0' means that we go over the entire punctured neighbourhoods U(0)\{0} , and not just their portions that are greater or less than 0.
Good old math yay! I simply said "negative infinity" since I only considered x approaching 1 as x = 0.99999999, but yeah it could have been x = 1.000000001 too.
Getting heavy Mean Girl vibes here...
My confusion is when you evaluate 1+. My understanding of “limit as X -> 1,” is that X gets very very close, but not quite 1. I’m confused why X would go past the limit.
look at the boxes of black and red pens on the right.
Im jealous too.
0+ and 0-
There is no limit
This is incredible. I cannot believe it. This has got to be some kind of a record. A miracle. I tried it and I got it right. I never get things right. This got to be the exception to prove the rule. Wait...
I once drew a piecewise function for my learners, below 2 it was defined as x and from 2 and above it was defined as x+2, I asked them to determine the limit of the function as it approaches 2. I realized they had long forgotten about approaching a point from the left and right, they've gotten used to simplifying then "substituting" the limit value.
in my opinion, this question is the best way to tell us why concepts are important.
This problem can only be solved by making the 1 a 1+ or 1-, giving us a way to represent limits going toward infinity.
Without the qualifier of + or - to show which infinity we are going toward, the problem doen't even exist. 1/1-0 is literally going nowhere because there are no directions to take.
1/1-0 is the mathematician's Schrödinger's Cat.
'This problem can only be solved by making the 1 a 1+ or 1-'
'Without the qualifier of + or - to show which infinity we are going toward, the problem doen't even exist'
No, this problem is perfectly solvable. Just watch the video.
Also, the expressions 'x->x_0+' and 'x->x_0-' do not refer to the tendency of something else towards any particular points at infinity. Those expressions simply tell us what portions of neighbourhoods of x_0 we are considering.
Im thankful to my teacher for giving importance to the concept of approaching from left and right side otherwise I couldn't have solved it
I got 0 by taking the reciprocal (1-1/x)^-1
Class need solve Demidovich and read Zorich
I'd add Engelking's 'General Topology' as a companion to Zorich, as Zorich takes an implicit topological approach in his textbooks.
We can get 0 as the solution by multiplying x-1 at both numerator and denominator
f(x) = x/(x- 1) = 1+ 1/(x-1)
let's consider 1/x and 1/(x-1)
A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift
1/(x-1) is a horizontal shift of the function 1/x, the graph will shift right
The simplest shift is a vertical shift, moving the graph up because this transformation involves adding a positive constant to the function 1/(x-1).
How the hell his whole class wasnt able to solve it, it's literally the basic LHL=RHL