@@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ Nope. sin (x) stays in the interval [-1, 1], and anything staying in a closed interval (meaning it's finite) over something going to infinity will go to zero. It's different when x goes to 0.
@@kevinm1317 sin x is close to x near 0. We are considering a limit going to infinity, where sin x is dominated by x. i.e. sin x = o(x). Approching infinity we have x/(x + sin x) = x /(x + o(x)) = 1 + o(1) o(1) tends to 0 by definition, hence the answer
I dont study math, but in highschool we were taught that if you have inf/inf conundrum, then you just bring out the highest power of x that you can, ie in this case x/(x+sinx) = x*(1)/(x(1+(sinx)/x) where the x-s cancel, leaving you with 1/(1+(sinx/x)) and as x-->inf sinx/x goes to 0, therefore you have 1/(1+0) = 1. Probably not the best solution, but it gets the answer in this case
@@zunaidparker hmmm, how is that "better" ? I think the squeeze theorem, in addition to providing the correct anwser, offers a very good glimpse at what is going on here.
But unless you are taking the fact as a given that sinx/x approaches 0 as x approaches infinity, you gotta prove exactly that with the squeeze theorem anyways
@@ianfowler9340 i get what youre saying but honestly i view the damped vibration as a private example of squeeze theorem anyways. you can also say that sinx/x is always between -1/x and 1/x and both approach 0 when x approaches infinity. but i guess its a matter of definition.
Worth noting that in the statement of L'hopitals rule, the requirement that the limit of f'/g' exists also includes the case that it exists but diverges, which is not typically counted as a limit existing. It only breaks down specifically at the case of an indeterminate limit
I'd just divide top and bottom by x and then it's pretty trivial. That's definitely a legal move, and then from that point you only need the limit of sin(x)/x as x approaches infinity. Idk how you approach that formally, but intuitively that's clearly 0, as sin is bounded between -1 and 1, so the magnitude of sin(x)/x decreases indefinitely as x increases. Edit: Apparently the squeeze theorem is the formal approach to such a limit.
For the sinx/x limit u can formally prove it using the sandwich/squeeze theorem. Since -1/x ≤ sinx/x ≤ 1/x and both the left and right side approach 0, sinx/x must also approach 0
@@sujals7108 You can prove it directly using a geometrical argument also. Just draw it and assume theta is smaller than pi/2 and greater than -pi/2, and the limit can be argumented without the sandwich theorem.
@harrabiwassim Sin is cyclical so it will never approach a value that doesn't appear in that range and will allows produce those values throughout infinity during every cycle
Which is probably what most people would do. It's sufficient that sin(x) is assimptotically smaller than x, the fact that it is bounded makes it even easier. I don't know why anyone's first instinct would be to use LH here
I solved it by inspection. I noticed that as x approaches infinity, x will be excessively bigger than sinx because sinx will be confined to the interval of -1
@@myself0510 1/(1+cosx) sort of swoops in from infinity, levels out to y=1/2, and then zips back to infinity repeatedly. as x goes to infinity, it never approaches any particular value, so the limit doesn’t exist and L'Hôpital's rule fails.
@@myself0510 denominator differentiates to 1-cosx, which doesn't converge as x tends to infinity (as cos oscillates). So the limit doesn't exist. A better approach is the squeeze theorem
I found very interesting that no comment here mentioned a critical part of LH often overlooked. The rule is technically defined for f(x)/g(c) with f,g: I ->R where I is an open interval - In this case any open intervall (a,infty) with a being any real number here. The function g, or better g’, has to satisfy g’(x) not equal to 0 for all x in that interval. This is not given here, as g’(x)=1+cos(x) has infinite zeroes in any interval (a,infty) you chose (because cos(x) takes the value -1 infinitely many times). Hence, LH can’t be applied here.
All of what you say is pointless. LH says that if (in addition to the other assumptions) the limit f'(x)/g'(x) exists, then the limit f(x)/g(x) also exists, and it is equal to the former limit. It does not say anything if the limit f'(x)/g'(x) does not exist. This is quite obvious if you go through the proof of LH in case both f and g have zero limits, and the limit is taken at a finite point, since the proof uses the Cauchy mean-value theorem.
I am not 100% sure, but I think you are mistaken and it's just that g(x) may not be equal to 0 for all x in (-infty, infty). So essentially it is not allowed to be g(x)=0.
@@DerKiesch I am not mistaken, but you are righjt for this reason. If g(x)=0 infinitely many times near the point where the limit is taken, then also g'(x)=0 infinitely many times by the mean-value theorem, and the limit f'(x)/g'x) does not exist (since it is meaningless near where the limit is taken). So, LH is inapplicable according to what I said. But there are many other reasons that the limit f'(x)/g'(x) may not exist, so singling out one specific reason is misleading.
This is a great point. Students sometimes tend to use the L'Hopital's rule without really checking the conditions. So, this is a great example to show the importance of checking the conditions to get a valid result. Otherwise someone would just argue that limit does not exist in this case, making an incorrect conclusion. Thank you for bringing this.
In my case (same with all other universities in Russia), L'Hopital's rule is mentioned, but only used one or two times. In calculus course it's used mostly to check if your solution is correct (which Wolfram does significantly better).
Yeah, I was wondering why you'd use L'Hopital here, it's simply the wrong tool. It makes about as much sense as bashing a screw into the wall with a hammer.
I think you can do a simpler proof than squeeze by just changing the expression to 1/(1+sinc(x)), where sinc(x) = sin(x)/x. Then you'll need squeeze if they don't know the limit x->infinity for sinc(x).
I love your “squeeze theorem” approach… Using the rigorous method of squinting I would have said 1, since it’s ♾️/(♾️+ [0,1]) = 1, but in a test you’d show something like what you did to cover yourself.
I would be crossed if showing that sin(x)/x -> 0 was not enough. Also, it is [-1,1] so half a point removed for that inaccuracy. And inf/inf+[-1,+1] is not always 1. 2x/(x+sin(x)) has a limit of 2.
How about an e and delta approach? Something like 1 - F(x) < e whenever x > d, where F(x) is 1/(1+sin(x)). In other words the difference between 1 and F(x) can be made arbitrarily small (less than any chosen arbitrary small number,e) with sufficiently large x > d is chosen, where d depends on e.
This is why I tell students to think in terms of growth rates, i.e., what we'd call asymptotics. For large x, the term x completely dominates sin(x) since sin(x) is bounded. So, we can visualize f(x) = x/(x + sin(x)) just "becoming" x/x as x gets larger and larger. The sine function just stops mattering. And this is true; we'd need to properly quantify "becoming" of course, but at the calc 1 level, it's good enough to say that sin(x) is just irrelevant compared to x for large x. And that logic is good enough to get pretty much all relevant limits.
Yeah, it's not hard if you don't think to use L'Hopital, but the subtlety about the exact statement of L'Hopital is something I haven't thought about in a long time. If the new ratio goes to a constant or exactly one of +/-infty, it still works, but this is an oscillating discontinuity, and the rule doesn't make a prediction.
@@uap4544 L'Hopital doesn't say what happens if f'/g' oscillates as x --> infty. (or as x --> c, or as x --> -infty) If f'/g' --> infty and f --> infty and g --> infty as x --> infty, then f/g --> infty. But f'/g' **oscillating** as x --> infty isn't one of the cases where the theorem comes to any conclusion whatsoever! It's a "dead end, go back one step and try something else" situation, because this situation doesn't even let you try to apply L'Hopital an additional time.
for the limit sin(x)/x as x->oo you need to use squeeze theorem either way This video is about an example of a limit that d'Hôpital's rule cannot solve
Just put x = 1/y. Then the limit becomes y -> 0 and you can just multiply the expression with y/y. The result is 1/(1 + ysin(1/y)). Now since sin is limited ysin(1/y) goes to 0 when y goes to 0. So it's just 1/(1+0) which is equal to 1.
The same method for finding the limit of f(x)/g(x) where the degree of f and g are equal works - divide above & below by x, and you get 1/(1+sin(x)/x)) - limit is obviously 1.
@@benheideveld4617 Because people will default to L'Hôpital. It's written in the thumbnail. The fact that the limit is clear by mere inspection is part of the cruelty.
Really nice work! I loved how you deconstructed the theorem. Couple thoughts: - Given a fraction, it is a common technique to try to find the "dominant" term among all terms in the numerator and denominator, and simplify by it. Clearly x, a function tending to infinity, dominates the bounded function sin(x). So just simplify by x as a knee-jerk reaction, and it will all be nice. - I finally learned the expression "squeeze theorem". Thanks! I never knew what it is called in English. In Hungary, we call it the "policeman law". The idea is that when two policemen grab a convict from each side, and walk into a cell, the convict is going to end up in the same cell. - If you wanna be really cruel, make it x^5/(x^5+sin(x)). Then the same problem only arises after five applications of L'Hôpital's rule. By that time, the students will definitely not think about where the hypotheses of the theorem ended and where the conclusion started. - Ah yeah, I think it is written as L'Hôpital's rule. So delete the "s", and put a hat on the "o". This is not so important, though.
Beautiful. A lot of the power of mathematics in science, for example, comes from the way math try to make all the assumptions explicit. This help. you find hidden biases and delimit the area of application (both expanding it, we need just this for it to work, and limiting it we need this for this to work). This is a beautiful remainder to pay attention.
What's devious here is that the indeterminate form of f/g is the correct type. But the f'/g' limit is 1/oscillating, which is **not** usable. If it were another 0/0 or infty/infty, we could try again (and maybe it works or maybe another dead end)
You could also add and substract "sin x" in the top part of the fraction, then break it into two. You get 1- sinx/(x+sin x) which goes to 1 as x goes to infinity
Just multiply by (1/x)/(1/x), then you get (x/x)/(x/x+sin(x)/x), simplify: 1/(1+sin(x)/x), and then as sin(x) oscillates between -1 and 1, it becomes insignificant as x goes to infinity, thus you get 1/1, which is just 1. Lol In most occasions where you have to solve a limit to infinity, multiply by 1 over the variable (to whichever power is needed, as in 1/x² or 1/x³), and it becomes a lot simpler, most stuff goes to zero and the rest is a lot easier to deal with.
Divide numerator and denomminaor by x 1/1+(sin(x)/x) replace x -> 1/y as x approaches inf, y approaches 0 So we have 1/1 + y sin(1/y) as sin is always between [-1,1] and y approaches 0 0* sin(1/y) = 0 We are left with 1/(1+0) = 1
I cheated and broke it out into (x/x)*(1/(1+sin(x)/x)). Going to infinity x/x is 1. The other factor 1/(1+sin(x)/x), sin(x)/x will go to 0 as sin(x) is always between (1,-1) and as x rises the term will become negligible. So you get (1/1) * (1/1+0)=1
This is why I like how analysis is thought in my uni. The powerful rules are left for the end, everything is derived from the mire basic rules (even stuff like limit of n^(1/n), big part of series is done without derivatives.
I sort of see why there can be two answers. The function becomes 1/(1+(sinx/x)), so 1. For x--> infinity, sinx/x goes to 0 and. it becomes 1/(1+0) = 1 ( I think sinx/x -->0 as x--> infinity uses the squeeze theorem which is what Michael used ultimately) 2. But, at x = 0 the sinx/x goes to 1 which gives 1/(1+1) = 1/2
Limsup and liminf are sneakily introduced (partially) as the squeeze theorem but then in elementary real analysis and measure theory they become a multitool that can do so much.
Easy if you you say the given limit is bounded between x/(x-1) and x/(x+1). Both bounds are equal to 1 in the limit, so the given limit problem is squeezed to equal 1.
It is bounded between x/(x+1) and x/(x-1), not between x/(x-1) and x/(x+1), which is probably the reason why he found it convenient to do it in a way that avoids that mistake.
im in 9th grade, i have a very basic undestanding of limits but i have never heard of the rules he talked about such as l'hospital or squeeze but bc the algebra and trig is extremely easy i fully understood! great teacher
Divide by x on top and bottom, use properties of limits and distribute the limit to each term. The limit as x approaches infinity of sin(x)/x is zero because you have a constant, finite but oscillating value getting divided by a ballooning infinitely large value. So it’s 1/(0 + 1) = 1
Oh, this should ABSOLUTELY be part of an early calculus class, because it perfectly demonstrates how to apply rules! Being strict is more important than anything, so the earlier an example like this is brought up, the better.
@@rafid.h.dejrah: First grammar not gramer. Second it has nothing to do with grammar which is the structure of sentences but with the meaning(s) of words. Look up pontificate and you'll see.
@@roberttelarket4934 I need to tell you that there are so many other ways of writing Grammar, and such related stuff... and for the vocab. it looks that u r not familiar with it plus u need to refresh ur english background info. mate.. methinks
After giving this some more thought, I tried to think of a new "squeeze theorem" that applies when x ----> +inf. -1 0 as x ---> +inf Therefore we are "squeezed" between 0 and 0. This may be the version some are referring to.
Write it as 1 - (sin(x)/(x + sin(x)). Notice how the top of the fraction bobs up and down whilst the magnitude of the bottom becomes as large as you please.
My Calc 3 teacher mentioned L'H in the same way that you did, and I thought he was making it needlessly complicated. And for just learning it, I think blindly plugging it in is probably the best way to introduce it. Because, "if it works, it works. But if it doesn't work, it doesn't work." is too complicated when just learning this weird rule.
The way we were taught, we're supposed to take the biggest term on top and bottom. Bigger on bottom means the limit is zero, bigger on top means the limit approaches infinity, but if the two terms have the same power then we divide to find the numerical answer.
In the example given, the derivative of the denominator at x = (2n+1)pi, where n is an integer, is 0; so the derivative of the numerator divided by the derivative of the denominator is undefined at values approaching infinity. A slightly better example to avoid this difficulty is Lim(x -> inf)(x + sin(x))/x .
L'Hopital rule gets a harsh treatment in Russian way of teaching analysis or just calculus. Despite its extreme simplicity and usefulness I guess you always start by transforming the expression or expanding functions at the point as power series or straight up using equivalents to infinitesimals. Like here you just can divide by x and that's it. I really liked the idea of using police theorem, though!
Iirc, it's exactly because it's a shortcut it's not used often. Goal is to teach and make students remember different methods, so if they have a problem L'Hopital's rule doesn't solve, they still know other methods
Same in Italy. Most of the time a power series expansion can solve the same kind of problems with undetermined forms and is easier to apply (De L'Hopital rules requires a few additiinal hypothesis checks ans sometimes recursive application). It's viewed mostly as a 'last resort' kind of technique.
You can intuitively figure out that this limit approaches 1 if you plug in bigger and bigger numbers, and you can tell the sin(x) bit plays a lesser and lesser role as x approaches infinity. Proving that is a different task.
The best and fast approach also called engineering approach in order to resolve this limit is bind x with a big number for instance 10000 and calculate the value of limit for this big number, and then you round the result.
My calc teacher taught me in cases like this to take only the most “powerful” term into account because everything else will be insignificant as you approach infinity. That leaves us with x/x=1
I'm pretty sure (unless there is a rule forbidding it, that I'm not aware of) it can be done much easier by getting taking x from nominator and denominator. Nominator: x*(1) Denominator: x*(1+sin(x)/x)) We know x is not 0 as it approaches infinity, so we can get rid of it, and we are lift with: 1/(1+sin(x)/x) when x is approaching infinity, sin(x)/x is approaching 0, so we can get rid of it as well. In the end, we are left with: 1/1 = 1
Since the sine of anything is always between --1 and +1, x/(x + sin(x)) will always be between x/(x+1) and x/(x-1). As x increases without limit, the +/--1 part becomes negligible and both x+1 and x--1 approach x. So the whole fraction approaches x/x = 1. L'Hopital's rule is not needed. The logic is similar to Rudin's famous function x sin(1/x), which is always between --x and +x. So as x approaches zero, x sin(1/x) also approaches zero, even though it crosses the x-axis infinitely many times between zero and 1/pi.
You just need to factor the denominator by "x" and then you can remove the "x" from the numerator. We end up with this: x / [x*(1+(sin(x)/x)] = 1 / [1+(sin(x)/x)] Furthermore, as x tends to infinity sin(x)/x = 0 The final result becomes, therefore, obvious: 1 / [1+(sin(x)/x)] = 1 / (1 + 0) = 1
From an intuitive perspective, as x ---> infinity, sin(x) alternates between --1 and 1, with both (+) & (--) halves of the graph symmetrical, so it's average value = 0.
just use the trick of adding and subtracting sinx in the numerator. So (x+sinx-sinx)/(x+sinx) can be written such 1+sinx/(sinx+x) and since sinx only takes values between -1 and 1 and the denominator is growing toward infinity the answer will just be 1+0=1
I just divided numerator and denominator by x, so the limit becomes lim (1/(1+(sin x)/x)) . But lim (sin x)/x when x goes to infinity is 0, since sin x is bounded and x is not. So the entire limit is 1
it’s easier to simplify by x on the numerator and denominator, you will have 1/(1+sinx/x), and we know sinx/x is 0, so you get your final result much quicker
when you approach a limit like this, because all values of sinx are bounded above and below, sinx can be treated like a constant. Then its pretty easy to see inf/(inf + a constant) = 1, assuming of course those infinities are the same, which in this case they are.
I just divided num and den by 1/x so you end up with Lim as x aproaches inf of 1/(1+sin(x)/x) Since sin(x)/x aproaches to zero, the limit aproaches to 1.
should tend to 1. the range of sin(x) as we know is between -1 and 1. so at infinity, sin(x) must equal some constant a between -1 and 1. we can rewrite the limit as lim [x/(x + a)] x -> inf at infinity, constants can be ignored as they do not affect the end behavior of the function. thus the limit becomes lim (x/x) x -> inf which just equals one.
My approach: Rewrite x as x + sin(x) - sin(x). Then the fraction becomes (x + sin(x) - sin(x))/(x + sin(x)) = 1 - sin(x)/(x + sin(x)) Limit of 1 is just 1, leaving only sin(x)/(x + sin(x)). Note that 1/(1-x)
2:55 why simply not say cos x = -1 0 +1 or something in between. You can clearly see that as x increases the thing swings from 1/2 to +inf. at the same time, x/x+sinx clearly converges to 1, because the swings are bounded and both bounds converge to 1. you don't need 10 minutes for that. I watched the rest and it's mostly about hopital. and this only applies to C1 class of functions, which is extremely rare.
the rate at which sin(x) oscillates does NOT change throughout the function (since it is 1 * sin(x) instead of x * sin(x) or something), so while x increases, y on average becomes closer and closer to 1 as the function strays closer and closer to x/x… it’s just like how 3/3.5 is much farther away from 1 than 1000000/1000000.5
I would do it by saying that the sinx is completely negligible since we’re dealing with infinities and the sinx only oscillates between -1 and 1, so you can rewrite it as the limit of x/x which is 1
My professor did not teach us LHopitals at ALL until after we learned chain rules and had our first dive into related rates. Glad we did though since he explained the dangers of mindlessly applying LHopitals to a lot of functions
We never learned it at all. It was only mentioned in some problems that said not to use it. It's just a shortcut and not a particularly useful one. I use Taylor expansions, which is essentially equivalent mathematically but comes with much more intuition.
Besides, L'Hôpital is so strong you can get used to it and you don't really think when you use that theorem, it's like doing an integral with a calculator.
As |sin(x)| >= 1 for real x, for all x>1: x/(x+1) =< x/(x+sin(x)) =< x/(x-1) Both x/(x-1) and x/(x+1) trivially go to 1 as x goes to infinity, so the limit is 1.
A much easier way would be to divide everything by x. Take limits and you are done. lim (1 / (1 + (sin(x) / x)) obviously goes to 1 as x increases without bound.
The limit 1/(1+cos x) does not exist because the numerator tends to 1 (it is constant regardless how large x gets) and the denominator oscillates forever between 0 and 2 as x grows (it is 1 + something that socilates between -1 and 1) so it cannot converge to anything. It is the same reason why the lim [x->inf] sin x does not exist. It is quite obvious and simple. No need for "sequential limits".
Just divide both numerator and denominator by x and solve directly for the limit: {1/(1 + sin(x) / x ) where x goes to infinity} We know that limit of sin(x)/x is zero, so the whole answer is going to be 1.
Dividing numerator and denominator by x yields 1/(1+sinx/x), both the new numerator and denominator go to 1 as x goes to infinity, so the limit is 1
Then you'd have to show that sin x/x -> 0 when x -> infty; also by the squeeze theorem.
@@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ it's quite known as a result and it's not so hard to prove
Right. This is just an abuse of the squeeze theorem. You’d have to justify that lim f(x) = lim 1/x f(x)
@@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ Nope. sin (x) stays in the interval [-1, 1], and anything staying in a closed interval (meaning it's finite) over something going to infinity will go to zero. It's different when x goes to 0.
@@florisv559 Right. I don't know what I was thinking when writing that lol.
I would have used the engineering approach. x >> sin(x) so sin(x) can be ignored
Also computer science approach lol
@@TheEternalVortex42 Also the right approach ;)
But the engineering approach tells you x=sin x so it should be 1/2
@@kevinm1317 sin x is close to x near 0. We are considering a limit going to infinity, where sin x is dominated by x. i.e. sin x = o(x).
Approching infinity we have
x/(x + sin x) = x /(x + o(x)) = 1 + o(1)
o(1) tends to 0 by definition, hence the answer
And physics approach. This is the case where intuition works much better than the heavy machinery of L'Hospital.
I dont study math, but in highschool we were taught that if you have inf/inf conundrum, then you just bring out the highest power of x that you can, ie in this case x/(x+sinx) = x*(1)/(x(1+(sinx)/x) where the x-s cancel, leaving you with 1/(1+(sinx/x)) and as x-->inf sinx/x goes to 0, therefore you have 1/(1+0) = 1. Probably not the best solution, but it gets the answer in this case
It is the "best" solution for this problem since it doesn't invoke any other theorems like the Squeeze Theorem.
@@zunaidparker hmmm, how is that "better" ? I think the squeeze theorem, in addition to providing the correct anwser, offers a very good glimpse at what is going on here.
This is probably the most obvious solution, it would still need squeeze to prove that the limit of sinx/x is zero in order to be complete, I think.
@@TimL_ yes true, I kinda overlooked that part but you're 100% right about that.
@@zunaidparker Except it does. Implicitly. You will prove that sin(x)/x→0 as x→∞ via squeeze theorem.
Just divide through by x
1/(1 + (sin(x)/x)) gives 1/(1+0) = 1
This. I considered this a high school level problem, not university level, tbh.
Explaining why l'Hospital fails is a good reminder, though.
But unless you are taking the fact as a given that sinx/x approaches 0 as x approaches infinity, you gotta prove exactly that with the squeeze theorem anyways
What about when your "x" is zero? Dividing by zero is excluded within the "definition of division" which is why we say "it's undefined."
@@MrSimmies x goes to infinity
@@ianfowler9340 i get what youre saying but honestly i view the damped vibration as a private example of squeeze theorem anyways. you can also say that sinx/x is always between -1/x and 1/x and both approach 0 when x approaches infinity. but i guess its a matter of definition.
Since sin x is bounded and x tends towards infinity, the dominant term of x+sin x is x. Therefore, x/(x+sin x)≈x/x=1.
valid aproach.
the video is about why L'H doesn't work
@@opensocietyenjoyer Yes, what‘s your point?
@@GrandAdmiralMitthrawnuruodo the point is that if you had watched the video, you would know that the comment you made misses the point
Worth noting that in the statement of L'hopitals rule, the requirement that the limit of f'/g' exists also includes the case that it exists but diverges, which is not typically counted as a limit existing. It only breaks down specifically at the case of an indeterminate limit
Yup, just checked it.
Very nice example. Some students have a way of memorizing theorems that emphasize conclusions, but not hypotheses. Examples like this help.
I'd just divide top and bottom by x and then it's pretty trivial. That's definitely a legal move, and then from that point you only need the limit of sin(x)/x as x approaches infinity. Idk how you approach that formally, but intuitively that's clearly 0, as sin is bounded between -1 and 1, so the magnitude of sin(x)/x decreases indefinitely as x increases.
Edit: Apparently the squeeze theorem is the formal approach to such a limit.
For the sinx/x limit u can formally prove it using the sandwich/squeeze theorem.
Since -1/x ≤ sinx/x ≤ 1/x and both the left and right side approach 0, sinx/x must also approach 0
@@sujals7108 Ah, cool - thanks. I've never used the squeeze theorem before but that was simple enough at least.
@@sujals7108 You can prove it directly using a geometrical argument also. Just draw it and assume theta is smaller than pi/2 and greater than -pi/2, and the limit can be argumented without the sandwich theorem.
@@andychow5509 sure... except theta is not smaller than pi/2 and greater than -pi/2? it's going to infinity.
@harrabiwassim Sin is cyclical so it will never approach a value that doesn't appear in that range and will allows produce those values throughout infinity during every cycle
I did it with the squeeze theorem before watching the video. I used the fact that x/(x+1)
Same
Which is probably what most people would do. It's sufficient that sin(x) is assimptotically smaller than x, the fact that it is bounded makes it even easier. I don't know why anyone's first instinct would be to use LH here
I did it by rearranging it into 1-sin x/(x+sin x) and got obvious solution
I solved it by inspection. I noticed that as x approaches infinity, x will be excessively bigger than sinx because sinx will be confined to the interval of -1
One good aspect of the squeeze or sandwich theorem is that it can prove the limit exists.
WRONG
That was my first thought as well.
@@firstname4337 Care to prove?
@@Handlessuck1 I'm guessing it's because inf/inf is indeterminate, but Idk, I'm new to calculus
I found myself totally engaged for this entire video. This was a delight
I remember learning Squeeze in 1st sem. calc, and L'Hôpital"s in 2nd; however, L'Hôpital's rule fast becomes a crutch. This example is delightful!!!
Just tried L'H and it doesn't work! I'm so amazed... Guess I need to go back and read the theorem again. Why doesn't it work in this case?
@@myself0510 1/(1+cosx) sort of swoops in from infinity, levels out to y=1/2, and then zips back to infinity repeatedly. as x goes to infinity, it never approaches any particular value, so the limit doesn’t exist and L'Hôpital's rule fails.
@@myself0510 denominator differentiates to 1-cosx, which doesn't converge as x tends to infinity (as cos oscillates). So the limit doesn't exist.
A better approach is the squeeze theorem
@@myself0510L'Hôpital's rule doesn't guarantee evaluability. All it gives you is a new form of the limit.
I found very interesting that no comment here mentioned a critical part of LH often overlooked. The rule is technically defined for f(x)/g(c) with f,g: I ->R where I is an open interval - In this case any open intervall (a,infty) with a being any real number here. The function g, or better g’, has to satisfy g’(x) not equal to 0 for all x in that interval. This is not given here, as g’(x)=1+cos(x) has infinite zeroes in any interval (a,infty) you chose (because cos(x) takes the value -1 infinitely many times). Hence, LH can’t be applied here.
All of what you say is pointless. LH says that if (in addition to the other assumptions) the limit f'(x)/g'(x) exists, then the limit f(x)/g(x) also exists, and it is equal to the former limit. It does not say anything if the limit f'(x)/g'(x) does not exist. This is quite obvious if you go through the proof of LH in case both f and g have zero limits, and the limit is taken at a finite point, since the proof uses the Cauchy mean-value theorem.
I am not 100% sure, but I think you are mistaken and it's just that g(x) may not be equal to 0 for all x in (-infty, infty). So essentially it is not allowed to be g(x)=0.
@@DerKiesch I am not mistaken, but you are righjt for this reason. If g(x)=0 infinitely many times near the point where the limit is taken, then also g'(x)=0 infinitely many times by the mean-value theorem, and the limit f'(x)/g'x) does not exist (since it is meaningless near where the limit is taken). So, LH is inapplicable according to what I said. But there are many other reasons that the limit f'(x)/g'(x) may not exist, so singling out one specific reason is misleading.
Just to underline the point I made, LH is also inapplicable when calculating the limit (x+sin x)/x at infinity, yet here g(x) is not 0 for positive x.
I mean your point nullifies if the function were x / (x+ 0.5*sin(x) );
LH is simply different from what you stated.
This is a great point. Students sometimes tend to use the L'Hopital's rule without really checking the conditions. So, this is a great example to show the importance of checking the conditions to get a valid result. Otherwise someone would just argue that limit does not exist in this case, making an incorrect conclusion. Thank you for bringing this.
In my case (same with all other universities in Russia), L'Hopital's rule is mentioned, but only used one or two times. In calculus course it's used mostly to check if your solution is correct (which Wolfram does significantly better).
Yeah, I was wondering why you'd use L'Hopital here, it's simply the wrong tool. It makes about as much sense as bashing a screw into the wall with a hammer.
I think you can do a simpler proof than squeeze by just changing the expression to 1/(1+sinc(x)), where sinc(x) = sin(x)/x. Then you'll need squeeze if they don't know the limit x->infinity for sinc(x).
I did it quite the same, but taking the inverse of the limit: lim (x+sin(x))/x = lim (1 + sin(x)/x) = 1.
I'm not sure why this is simpler, seeing as this requires knowing about the limits of sinc(x) which you'd already prove using the squeeze theorem
@@epicmarschmallow5049 The squeeze theorem for sin(x)/x is simpler, because it's bounded by ± 1/x.
I love your “squeeze theorem” approach…
Using the rigorous method of squinting I would have said 1, since it’s ♾️/(♾️+ [0,1]) = 1, but in a test you’d show something like what you did to cover yourself.
I would be crossed if showing that sin(x)/x -> 0 was not enough. Also, it is [-1,1] so half a point removed for that inaccuracy. And inf/inf+[-1,+1] is not always 1.
2x/(x+sin(x)) has a limit of 2.
@@57thorns it's more like 1/(1+[-1,1]/infinity). this is 1.
How about an e and delta approach? Something like 1 - F(x) < e whenever x > d, where F(x) is 1/(1+sin(x)). In other words the difference between 1 and F(x) can be made arbitrarily small (less than any chosen arbitrary small number,e) with sufficiently large x > d is chosen, where d depends on e.
This is why I tell students to think in terms of growth rates, i.e., what we'd call asymptotics. For large x, the term x completely dominates sin(x) since sin(x) is bounded. So, we can visualize f(x) = x/(x + sin(x)) just "becoming" x/x as x gets larger and larger. The sine function just stops mattering. And this is true; we'd need to properly quantify "becoming" of course, but at the calc 1 level, it's good enough to say that sin(x) is just irrelevant compared to x for large x. And that logic is good enough to get pretty much all relevant limits.
Yeah, it's not hard if you don't think to use L'Hopital, but the subtlety about the exact statement of L'Hopital is something I haven't thought about in a long time. If the new ratio goes to a constant or exactly one of +/-infty, it still works, but this is an oscillating discontinuity, and the rule doesn't make a prediction.
don't remember but i wasn't only continuity can be applied l'hopital? or a simplification of the past... :)
@@uap4544 L'Hopital doesn't say what happens if f'/g' oscillates as x --> infty. (or as x --> c, or as x --> -infty)
If f'/g' --> infty and f --> infty and g --> infty as x --> infty, then f/g --> infty. But f'/g' **oscillating** as x --> infty isn't one of the cases where the theorem comes to any conclusion whatsoever! It's a "dead end, go back one step and try something else" situation, because this situation doesn't even let you try to apply L'Hopital an additional time.
this is pointlessly complicated, just factor out x from numerator and denominator and compute the limit of them both to be 1.
for the limit sin(x)/x as x->oo you need to use squeeze theorem either way
This video is about an example of a limit that d'Hôpital's rule cannot solve
Yeah but where do you get that limit sinx/x is 1? The squeeze theorem
@@hydropage2855limit sin x/x is zero as x goes to infinity.
The point is to provide an actual counterexample to L'Hôpital's rule.
9:41 I like how the conclusion was not gauranteed :p
divide both denominator and numerator by x, in numerator we get sinx/x and x tends to inf. while sinx can go max to 1 it equal to 0. so 1/1+0=1
Just put x = 1/y. Then the limit becomes y -> 0 and you can just multiply the expression with y/y. The result is 1/(1 + ysin(1/y)). Now since sin is limited ysin(1/y) goes to 0 when y goes to 0. So it's just 1/(1+0) which is equal to 1.
The same method for finding the limit of f(x)/g(x) where the degree of f and g are equal works - divide above & below by x, and you get 1/(1+sin(x)/x)) - limit is obviously 1.
x/(x+sinx)=1/(1+sincx)
Lim as x goes to infinity of sinc x is 0.
Therefore the limit is 1/(1+0)=1
Exactly, why is this cruel? Or did he mean crude?
I have learned something NEW here.
@@benheideveld4617 Because people will default to L'Hôpital. It's written in the thumbnail. The fact that the limit is clear by mere inspection is part of the cruelty.
@@benheideveld4617 It's not cruel.
Really nice work! I loved how you deconstructed the theorem. Couple thoughts:
- Given a fraction, it is a common technique to try to find the "dominant" term among all terms in the numerator and denominator, and simplify by it. Clearly x, a function tending to infinity, dominates the bounded function sin(x). So just simplify by x as a knee-jerk reaction, and it will all be nice.
- I finally learned the expression "squeeze theorem". Thanks! I never knew what it is called in English. In Hungary, we call it the "policeman law". The idea is that when two policemen grab a convict from each side, and walk into a cell, the convict is going to end up in the same cell.
- If you wanna be really cruel, make it x^5/(x^5+sin(x)). Then the same problem only arises after five applications of L'Hôpital's rule. By that time, the students will definitely not think about where the hypotheses of the theorem ended and where the conclusion started.
- Ah yeah, I think it is written as L'Hôpital's rule. So delete the "s", and put a hat on the "o". This is not so important, though.
In Poland we call it "thoerem about three sequences", it is so simple and a little bit uncreative compared to other languages
L'Hôpital and L'Hospital are the same. In French, ô is shorthand for os, and I don't know why.
Also goddamn its called policeman law in Hungary???
just for the sake of curiosity, in italy we call it "teorema dei due carabinieri" for the exact same reason and students find this thing hilarious 😂
In portuguese we call it the sandwich theorem hahaha pretty self explanatory
Beautiful. A lot of the power of mathematics in science, for example, comes from the way math try to make all the assumptions explicit. This help. you find hidden biases and delimit the area of application (both expanding it, we need just this for it to work, and limiting it we need this for this to work). This is a beautiful remainder to pay attention.
divide both nominator and denominator by x to get 1/(1+ sinx/x) and use sinx/x -> 0 to show that lim f = 1
I loved this problem as it is a good reminder to consider all the precise requirements for a rule/theorem before applying it.
Which is why I avoid using these more complex rules, if possible.
What's devious here is that the indeterminate form of f/g is the correct type. But the f'/g' limit is 1/oscillating, which is **not** usable. If it were another 0/0 or infty/infty, we could try again (and maybe it works or maybe another dead end)
Let x = 1/t so
Lt. 1/1+tsin(1/t)
t-->0+
Since sin of any real number is between -1 and 1, tsin(1/t)=0, so limit is equal to 1
You could also add and substract "sin x" in the top part of the fraction, then break it into two. You get 1- sinx/(x+sin x) which goes to 1 as x goes to infinity
Let x=1/t so when x->inf, t->0 and the function is (1/t)/((1/t)+Sin(1/t)). Solving lim:t-->0 1/(1+t*Sin(1/t)) = 1/(1+0) = 1
Just multiply by (1/x)/(1/x), then you get (x/x)/(x/x+sin(x)/x), simplify: 1/(1+sin(x)/x), and then as sin(x) oscillates between -1 and 1, it becomes insignificant as x goes to infinity, thus you get 1/1, which is just 1. Lol
In most occasions where you have to solve a limit to infinity, multiply by 1 over the variable (to whichever power is needed, as in 1/x² or 1/x³), and it becomes a lot simpler, most stuff goes to zero and the rest is a lot easier to deal with.
Divide numerator and denomminaor by x
1/1+(sin(x)/x)
replace x -> 1/y
as x approaches inf, y approaches 0
So we have
1/1 + y sin(1/y)
as sin is always between [-1,1] and y approaches 0
0* sin(1/y) = 0
We are left with 1/(1+0) = 1
I cheated and broke it out into (x/x)*(1/(1+sin(x)/x)). Going to infinity x/x is 1. The other factor 1/(1+sin(x)/x), sin(x)/x will go to 0 as sin(x) is always between (1,-1) and as x rises the term will become negligible. So you get (1/1) * (1/1+0)=1
This is why I like how analysis is thought in my uni. The powerful rules are left for the end, everything is derived from the mire basic rules (even stuff like limit of n^(1/n), big part of series is done without derivatives.
I sort of see why there can be two answers. The function becomes 1/(1+(sinx/x)), so
1. For x--> infinity, sinx/x goes to 0 and. it becomes 1/(1+0) = 1 (
I think sinx/x -->0 as x--> infinity uses the squeeze theorem which is what Michael used ultimately)
2. But, at x = 0 the sinx/x goes to 1 which gives 1/(1+1) = 1/2
-1
5:37 also notice that even if x is negative, the process is still exactly the same, so one could even take both cases into consideration
We can also replace x with 1/y
as x -> inf
y -> 0
and we get the limit equal to 1.
Limsup and liminf are sneakily introduced (partially) as the squeeze theorem but then in elementary real analysis and measure theory they become a multitool that can do so much.
It would be nice to show the logic behind why L'Hospital's Rule normally works, and why it breaks down if a limit doesn't exist.
Because it depends on the mean value theorem.
We can use L'Hopital's rule by noting that for x > 0, x / (x + 1)
Easy if you you say the given limit is bounded between x/(x-1) and x/(x+1). Both bounds are equal to 1 in the limit, so the given limit problem is squeezed to equal 1.
It is bounded between x/(x+1) and x/(x-1), not between x/(x-1) and x/(x+1), which is probably the reason why he found it convenient to do it in a way that avoids that mistake.
@@pedroteran5885 It's bounded between yet not bounded between the same two functions?
im in 9th grade, i have a very basic undestanding of limits but i have never heard of the rules he talked about such as l'hospital or squeeze but bc the algebra and trig is extremely easy i fully understood! great teacher
Divide by x on top and bottom, use properties of limits and distribute the limit to each term.
The limit as x approaches infinity of sin(x)/x is zero because you have a constant, finite but oscillating value getting divided by a ballooning infinitely large value.
So it’s 1/(0 + 1) = 1
Oh, this should ABSOLUTELY be part of an early calculus class, because it perfectly demonstrates how to apply rules! Being strict is more important than anything, so the earlier an example like this is brought up, the better.
At first glance, I see that -1
One of the best channels that is really working on the pontification of math trajectory
Pontification is the wrong word. It has a negative connotation which you I assume you didn't imply here!
@@roberttelarket4934 well, according to the old English Gramer, it is very commonly used for Unique stuff...
@@rafid.h.dejrah: First grammar not gramer. Second it has nothing to do with grammar which is the structure of sentences but with the meaning(s) of words. Look up pontificate and you'll see.
@@roberttelarket4934 I need to tell you that there are so many other ways of writing Grammar, and such related stuff... and for the vocab. it looks that u r not familiar with it plus u need to refresh ur english background info. mate.. methinks
@@rafid.h.dejrah: I was born and live in the U.S. and my knowledge of English is more than sufficient!
Maybe squeeze theorem is method of choice
After giving this some more thought, I tried to think of a new "squeeze theorem" that applies when x ----> +inf.
-1 0 as x ---> +inf
Therefore we are "squeezed" between 0 and 0.
This may be the version some are referring to.
Write it as 1 - (sin(x)/(x + sin(x)). Notice how the top of the fraction bobs up and down whilst the magnitude of the bottom becomes as large as you please.
My Calc 3 teacher mentioned L'H in the same way that you did, and I thought he was making it needlessly complicated. And for just learning it, I think blindly plugging it in is probably the best way to introduce it. Because, "if it works, it works. But if it doesn't work, it doesn't work." is too complicated when just learning this weird rule.
The fact that you people face Hôpitsl rule in calculus THREE only makes things more hilarious. America. Lol.
The way we were taught, we're supposed to take the biggest term on top and bottom. Bigger on bottom means the limit is zero, bigger on top means the limit approaches infinity, but if the two terms have the same power then we divide to find the numerical answer.
1/(1 + sin(x) / x) = 1 / (1 + 0) = 1. sin(x) / x -> 0 since -1
In the example given, the derivative of the denominator at x = (2n+1)pi, where n is an integer, is 0; so the derivative of the numerator divided by the derivative of the denominator is undefined at values approaching infinity. A slightly better example to avoid this difficulty is Lim(x -> inf)(x + sin(x))/x .
L'Hopital rule gets a harsh treatment in Russian way of teaching analysis or just calculus. Despite its extreme simplicity and usefulness I guess you always start by transforming the expression or expanding functions at the point as power series or straight up using equivalents to infinitesimals. Like here you just can divide by x and that's it.
I really liked the idea of using police theorem, though!
Iirc, it's exactly because it's a shortcut it's not used often. Goal is to teach and make students remember different methods, so if they have a problem L'Hopital's rule doesn't solve, they still know other methods
My teacher used to be an ass about limits... this may be why
Same in Italy.
Most of the time a power series expansion can solve the same kind of problems with undetermined forms and is easier to apply (De L'Hopital rules requires a few additiinal hypothesis checks ans sometimes recursive application).
It's viewed mostly as a 'last resort' kind of technique.
I would never use LH for a limit like this. x dominates sin x for x >= 6.28. The limit is therefor of the form x/(x + K) which tends to 1.
You can intuitively figure out that this limit approaches 1 if you plug in bigger and bigger numbers, and you can tell the sin(x) bit plays a lesser and lesser role as x approaches infinity. Proving that is a different task.
I can't see how anyone could not immediately know the answer if they know anything about limits.
The best and fast approach also called engineering approach in order to resolve this limit is bind x with a big number for instance 10000 and calculate the value of limit for this big number, and then you round the result.
My calc teacher taught me in cases like this to take only the most “powerful” term into account because everything else will be insignificant as you approach infinity. That leaves us with x/x=1
good reflex, but carefull with trigo function absolute limit theorem and squeeze theorem are usefool tools with those functions
Isn't there are very simple solution just by rewriting x /(x+sinx) as 1-sinx/(x+sinx) which is easy to see goes to 1
Shortcut: Divide through by x to instead get the limit of 1/(1 + sinc(x)) which is known to be 1.
My method.
squeeze theorem. It's a one-liner
I'm pretty sure (unless there is a rule forbidding it, that I'm not aware of) it can be done much easier by getting taking x from nominator and denominator.
Nominator: x*(1)
Denominator: x*(1+sin(x)/x))
We know x is not 0 as it approaches infinity, so we can get rid of it, and we are lift with:
1/(1+sin(x)/x)
when x is approaching infinity, sin(x)/x is approaching 0, so we can get rid of it as well.
In the end, we are left with:
1/1 = 1
Could also note that x/(x+sinx)=1-sinx/(x+sinx). Then -1 0 and we get 1 as desired.
The easiest for me was just adding and subtracting `sin x` at the top, then it's 1 - (sinx / x+sinx) < 1 - (sinx / x) = 0
But actually 1-sinx/(x+sinx)>=1-sinx/x
Just factor out x in numerator and denominator, then use sin(x)/x -> 0. Done.
Since the sine of anything is always between --1 and +1, x/(x + sin(x)) will always be between x/(x+1) and x/(x-1). As x increases without limit, the +/--1 part becomes negligible and both x+1 and x--1 approach x. So the whole fraction approaches x/x = 1. L'Hopital's rule is not needed.
The logic is similar to Rudin's famous function x sin(1/x), which is always between --x and +x. So as x approaches zero, x sin(1/x) also approaches zero, even though it crosses the x-axis infinitely many times between zero and 1/pi.
You just need to factor the denominator by "x" and then you can remove the "x" from the numerator.
We end up with this: x / [x*(1+(sin(x)/x)] = 1 / [1+(sin(x)/x)]
Furthermore, as x tends to infinity sin(x)/x = 0
The final result becomes, therefore, obvious:
1 / [1+(sin(x)/x)] = 1 / (1 + 0) = 1
From an intuitive perspective, as x ---> infinity, sin(x) alternates between --1 and 1, with both (+) & (--) halves of the graph symmetrical, so it's average value = 0.
Did anybody else see the broken piece of the bow hanging off the side?
That’s how you know it’s the person who’s holding it all together.
Good job
just use the trick of adding and subtracting sinx in the numerator. So (x+sinx-sinx)/(x+sinx) can be written such 1+sinx/(sinx+x) and since sinx only takes values between -1 and 1 and the denominator is growing toward infinity the answer will just be 1+0=1
I just divided numerator and denominator by x, so the limit becomes lim (1/(1+(sin x)/x)) . But lim (sin x)/x when x goes to infinity is 0, since sin x is bounded and x is not. So the entire limit is 1
Just divide by x and it's pretty straightforward after that
it’s easier to simplify by x on the numerator and denominator, you will have 1/(1+sinx/x), and we know sinx/x is 0, so you get your final result much quicker
Sir, the limit is visible here, just divide both the numerator and the denominator by x and then limit to infinity we can get 1／(1+（0／inf）） =1
-1
for people that are fast:
sinx = o(x) as x->inf, as it is at most 1 or -1 (Landau's Symbol). Thus, lim x/x+o(x)=x/x=1.
when you approach a limit like this, because all values of sinx are bounded above and below, sinx can be treated like a constant. Then its pretty easy to see inf/(inf + a constant) = 1, assuming of course those infinities are the same, which in this case they are.
I just divided num and den by 1/x so you end up with Lim as x aproaches inf of 1/(1+sin(x)/x) Since sin(x)/x aproaches to zero, the limit aproaches to 1.
For x > π:
|x / (x + sin x) -1| ≤ |x / (x - 1) -1| = |1/(x-1)| --> 0 for x --> ∞
So by definition of limit the limit is 1.
should tend to 1.
the range of sin(x) as we know is between -1 and 1. so at infinity, sin(x) must equal some constant a between -1 and 1. we can rewrite the limit as
lim [x/(x + a)]
x -> inf
at infinity, constants can be ignored as they do not affect the end behavior of the function. thus the limit becomes
lim (x/x)
x -> inf
which just equals one.
Since sin x will always fluctuate between [-1,1] and x tends to infinity,can we not ignore sin x as x>>>>>sin x. Our problem becomes x/x =1
My approach:
Rewrite x as x + sin(x) - sin(x).
Then the fraction becomes
(x + sin(x) - sin(x))/(x + sin(x))
= 1 - sin(x)/(x + sin(x))
Limit of 1 is just 1, leaving only sin(x)/(x + sin(x)).
Note that 1/(1-x)
sinx has rage of -1 to 1, which is negligible when limit x goes to inifnity so it is basically x/x = 1
2:55 why simply not say cos x = -1 0 +1 or something in between.
You can clearly see that as x increases the thing swings from 1/2 to +inf.
at the same time, x/x+sinx clearly converges to 1, because the swings are bounded and both bounds converge to 1.
you don't need 10 minutes for that.
I watched the rest and it's mostly about hopital. and this only applies to C1 class of functions, which is extremely rare.
From the beginning,you could have divided by x up and down, leaving you with 1/(1+sinx/x) ->1 since sinx/x->0
the rate at which sin(x) oscillates does NOT change throughout the function (since it is 1 * sin(x) instead of x * sin(x) or something), so while x increases, y on average becomes closer and closer to 1 as the function strays closer and closer to x/x… it’s just like how 3/3.5 is much farther away from 1 than 1000000/1000000.5
L'Hospital rule should always be the last resort.
I would do it by saying that the sinx is completely negligible since we’re dealing with infinities and the sinx only oscillates between -1 and 1, so you can rewrite it as the limit of x/x which is 1
My professor did not teach us LHopitals at ALL until after we learned chain rules and had our first dive into related rates. Glad we did though since he explained the dangers of mindlessly applying LHopitals to a lot of functions
We never learned it at all. It was only mentioned in some problems that said not to use it. It's just a shortcut and not a particularly useful one. I use Taylor expansions, which is essentially equivalent mathematically but comes with much more intuition.
Besides, L'Hôpital is so strong you can get used to it and you don't really think when you use that theorem, it's like doing an integral with a calculator.
Took me more to open the clip than to solve it. This is one of those instant limits. Try the diabolic limit, that one is fun!
As |sin(x)| >= 1 for real x, for all x>1:
x/(x+1) =< x/(x+sin(x)) =< x/(x-1)
Both x/(x-1) and x/(x+1) trivially go to 1 as x goes to infinity, so the limit is 1.
A much easier way would be to divide everything by x. Take limits and you are done.
lim (1 / (1 + (sin(x) / x)) obviously goes to 1 as x increases without bound.
The limit 1/(1+cos x) does not exist because the numerator tends to 1 (it is constant regardless how large x gets) and the denominator oscillates forever between 0 and 2 as x grows (it is 1 + something that socilates between -1 and 1) so it cannot converge to anything. It is the same reason why the lim [x->inf] sin x does not exist. It is quite obvious and simple. No need for "sequential limits".
as -1 < sin(x) < 1,
LIMx->inf x/(x+sin(x)) (sin(x) is negligible)
is equal to x/x = 1
Just divide both numerator and denominator by x and solve directly for the limit:
{1/(1 + sin(x) / x ) where x goes to infinity}
We know that limit of sin(x)/x is zero, so the whole answer is going to be 1.
Divide bith sides by x and then u only need to solve sinx/x which is 0 and the limit turns out to be 1