When I took Calc 3 my teacher had this as a test question: True/False If the limit exists along every straight path through a point then the limit exists at the point. (This was the final) Some unexplainable mathematical intuition told me that it had to be false I just couldn’t quite prove why, so I went with my gut and answered false. When I got my test back I had gotten everything right except for that question. I feel both a sense of intense satisfaction and anger from this video, if only I found it at the time to show the teacher!
@@ilias-4252 I mean.... both here aren't hard. The part where it is wrong is the "straight [path]" restriction, you must check for *every* path (straight and curved).... It's not hard to add inadvertently when remembering something(which the original commenter might have done)... or to add without making it incorrect(which commenter's teacher might have done)
At least getting all curved paths to agree is enough to prove the limit. If we work with 3-variable functions, do we need all "curved planes" to agree as well?
To clarify @@d.l.7416's answer further, note that the result when a limit does not exist (7:03) only specifies the existence some "sequence of points approaching zero" disagreeing with the limit L, meaning each axis/dimension sequence approaches 0 individually, and the sequence of the function as a whole disagrees with L. The number of dimensions doesn't matter here since nothing else about the sequence of points is specified, and you can simply draw an N-dimensional curved path through such a sequence.
@@magneticflux- Additionally, if every path of approach agrees, then every plane of approach also agrees, since any plane is composed of a set of paths
I used to be a Calculus II tutor at university and I had a similar "trick" to test all possible line paths that didn't need to check the case x=0 separately. All I did was x=at and y=bt and took the limit with t going to zero. This method could also be generalised to allow limits that go towards any generic point (x0,y0) by having x=at+x0 and y=bt+y0 with t going to zero. Most limits would end up depending on a or b, meaning they give different results for different paths, and this, do not exist.
delightful stuff! I saw the piecewise path coming, but you surprised me with the version you found 'in the wild.' does "all curved paths" imply smoothness (i.e., C_inf)? your board work is good, but could be leveled up by adding contrasting-color annotations to indicate transformations between lines.(also good practice in a lecture hall, but it can be hard to keep up with. time works differently here, though.) and at 6:40 you ran out of space and looped back around to the top, so by 6:47 the notation flow is ambiguous. you showed restraint in highlighting some failure modes of graphing calculators without dunking on them. I expect you know that you could graph look-alike replicas of your functions by offsetting them a tiny bit to shift the hole in the range onto a gap in the domain lattice that's used to sample and render the surface.
Thanks! I assume you mean "is it enough to check only smooth paths?" I think it depends on your coordinate system. If we're writing paths as y as a function of x then no, since x=0 isn't even a function. But in terms of polar coordinates, writing theta as a function of r, I'm pretty sure it is enough to check only smooth ones as long as you don't require smoothness at r=0. The function f(x)=e^(-1/(1-x^2)) is smooth on [-1,1] and all derivatives are 0 at -1 and 1, so if you have a sequence of points like at 7:02 you could piece together shifted and scaled copies of this to get a function theta(r) passing through the points.
You didn’t specify it at the very end, but checking every curved path only works if you keep yourself to rational functions. Otherwise you can keep playing piecewise games.
Unless I’m mistaken, checking every curved path should work for piecewise functions as well. For the limit to not equal some value L, there must be some value E such that there is a point within every radius D of (0,0) for which the function differs from L by at least E. Taking D to be 1/2, 1/3, 1/4 and so on gives an infinite sequence of points, and then you can draw a spiralling path through them. So if the limit doesn’t equal L then it must not equal L along some path, and conversely if the limit is equal on every path then the overall limit must be the same.
@@josephnewton The simplest counterexample to the general statement: let f(x, y) = 1 at (0, 0) and 0 elsewhere. Any line terminating at (0, 0) also works. For curves that pass through the origin, the only places where rational functions are nondifferentiable are at poles. So either your rational function is locally linear at the origin or it's a pole - either one would not fit general continuous piecewise functions. So I'm not sure that even that is a strong enough guarantee. (Not totally clear on this second paragraph though)
There's a difference between the limit at 0 existing and the function being continuous at 0; the piecewise example you provided is discontinuous, but the limit does exist (and equals 0).
@@wsshambaugh To add a bit to the other response, if you are looking at the limit as x aproaches a of f(x), the value f(a) doesn't matter(of course I implicitly assumed f is defined at a, but that doesn't matter either).
@@wsshambaugh Right, I forgot to include that the point within radius D can't be at (0,0). The full definition is that the limit at (x',y') equals L if for every E>0 there exists D>0 such that 0
Even if it can't calculate 0, that gap is huge. It's like it's ignoring anything below 1. If it's just 0 it ignores, then it should just be a really tiny point.
The calculator only samples a grid of points, which are the vertices of the green checkerboard (that’s why there’s exactly 4 squares missing in the middle). I could have made the gap smaller by increasing the number of vertices, or just got rid of it by having (0,0) be between the sampled points, but the later functions in the video are a lot harder to graph accurately because of the big height changes
Btw your “unnatural” counterexample also swiftly shows that existence of partial derivatives (they’re all 0 at the origin) does not imply differentiability (function is not even continuous at the origin, let alone differentiable). A comment regarding the “check all curved paths” thing - one may show that the limit of f at, say, the origin exists, iff the limits of a FINITE family of restrictions of f to a collection of DISJOINT subsets whose union is the domain of f (all having the origin as an accumulation point blah blah) all exist and coincide. This is blatantly false for infinite families of disjoint subsets, e.g. the collection of straight lines through the origin. However one may certainly consider an infinite family of non-overlapping CURVED paths that accumulate at the origin, along which the limit exists, and all of the limits coincide, but the overall limit still doesn’t exist. Take the unnatural counterexample again, and probe it along, say, curves of the form y = Ax^3 for real A, or logarithmic spirals that converge to 0… Point is, the issue isn’t straight vs curved, but disjoint vs overlapping! Indeed, the theorem becomes true again if you allow the subsets to overlap.
@@Rotem_S wouldn't the same counter example he first presented with a quadratic for straight lines work just fine if we sub out the quadratic for e.g. an exponential?
I was taught that a limit of infinity is not the same thing as does not exist. You only get DNE if one side is positive infinity and the other side is negative infinity. So, for example, the lim_x->0 (1/|x|) = infinity, but lim_x->0 (1/|x|) Does Not Exist.
Our math prof actually preferred the definition of the limit agreeing on all paths over the epsilon delta definition. Maybe I actually should have gone for math instead of compsi but oh well
Funny enough this is the kind of thing that Ted kaczynski wrote about in his thesis. The points you’re describing are called “ambiguity points” and there can only be countable many of them regardless of the function
Can a function, at a point where it *is* defined, have a limit not equal to the function at that point? Or is the limit of a function at every defined point by definition the value of the function at that point?
I'm confused, why can't that piecewise polynomial function have a limit of 0? It's been a couple years since calculus, but I don't seem to remember "if the function is defined at x, the limit is either equal to f(x) or does not exist" being a rule. Wouldn't f(x) = {0 if x = 0, 1 otherwise} have a limit of 0 as x goes to 0?
That’s correct, the limit can be different from the value of the function. But for the function f(x,y)=1 when x=y^2, there are points (x,y) arbitrarily close to (0,0) such that f(x,y)=1, so the limit can’t be zero
depends on how you do it, but often the proof of sinx/x -> 1 will be a *prerequisite* for finding d(sinx)/dx, which you need for both the Taylor series and L'Hôpital's rule
I don't understand why the limit along the straight lines equals 0. You can get arbitrarily close to 1 along any of the straight lines, so it should equal 1.
You can get arbitrarily close in _distance_ to the point where the value is 1, but not in value. On all of those increasingly closer points, the value is still 0 (except where you actually cross the parabola) and does not approach 1. Instead, it jumps to 1 abruptly when you reach the end point. (Assuming that you were talking about the function at 4:08.)
@@bscutajar Exactly :). The situation here is a bit more complicated than usual because there are two input variables and therefore infinitely many ways to move around the input plane. If you follow the function along the path of the parabola, or along a path which never crosses the parabola, the function is continuous for all points along that path. But if you consider the entire input plane (and all possible paths through it), the function is discontinuous.
I think x^y is a good example. y = mx gives (x^x)^m which approaches 1. x=0 gives 0^y, which gives 0. and x = A^(1/y) gives A. so most lines give you 1, one line gives you 0 and some curves give you any number you want. My question is what set of functions give you the same limit when you approach 0 with a curve as with a line tangent to the curve. Note that different curves are still allowed to give different limits.
2:00 Actually when using interchanged implication, it is not sufficient just negating the sides, they need to be negated and interchanged at once, either interchanged implication method of proof is not valid
I'm glad I'm not the only one who finds functions in the wild like this.
that's odd, all i seem to find in my local bush are dysfunctions
I only find epsilons
That can't be a good omen, variety is the spice of life. Although usually when I find a gamma it's a really bad sine.
I thought I hunted all the wild functions to extinction. Where do you find your wild functions now of days?
When I took Calc 3 my teacher had this as a test question: True/False If the limit exists along every straight path through a point then the limit exists at the point. (This was the final) Some unexplainable mathematical intuition told me that it had to be false I just couldn’t quite prove why, so I went with my gut and answered false. When I got my test back I had gotten everything right except for that question. I feel both a sense of intense satisfaction and anger from this video, if only I found it at the time to show the teacher!
Are you sure you aren't misremembering? Keep in mind that s normally more likely than your answer being correct and the teacher making a mistake.
@@ilias-4252 I mean.... both here aren't hard.
The part where it is wrong is the "straight [path]" restriction, you must check for *every* path (straight and curved)....
It's not hard to add inadvertently when remembering something(which the original commenter might have done)... or to add without making it incorrect(which commenter's teacher might have done)
@@floppathebased1492 Proving a statement like that is real analysis, not calculus 3.
This is exactly what I needed to see in calc 3, I never understood how a function like that could exist but you explained it perfectly!
At least getting all curved paths to agree is enough to prove the limit. If we work with 3-variable functions, do we need all "curved planes" to agree as well?
no you just need paths.
like in 1d you use a path, in 2d you use a path, and similarly in any dimension you use paths
To clarify @@d.l.7416's answer further, note that the result when a limit does not exist (7:03) only specifies the existence some "sequence of points approaching zero" disagreeing with the limit L, meaning each axis/dimension sequence approaches 0 individually, and the sequence of the function as a whole disagrees with L. The number of dimensions doesn't matter here since nothing else about the sequence of points is specified, and you can simply draw an N-dimensional curved path through such a sequence.
Right, a series of points is just a path, regardless of dimentionality. Thanks!
Is there any direct proof of it? Because a counter-example feels unsatisfactory even if the function looks good
@@magneticflux- Additionally, if every path of approach agrees, then every plane of approach also agrees, since any plane is composed of a set of paths
I used to be a Calculus II tutor at university and I had a similar "trick" to test all possible line paths that didn't need to check the case x=0 separately. All I did was x=at and y=bt and took the limit with t going to zero. This method could also be generalised to allow limits that go towards any generic point (x0,y0) by having x=at+x0 and y=bt+y0 with t going to zero. Most limits would end up depending on a or b, meaning they give different results for different paths, and this, do not exist.
delightful stuff! I saw the piecewise path coming, but you surprised me with the version you found 'in the wild.' does "all curved paths" imply smoothness (i.e., C_inf)?
your board work is good, but could be leveled up by adding contrasting-color annotations to indicate transformations between lines.(also good practice in a lecture hall, but it can be hard to keep up with. time works differently here, though.) and at 6:40 you ran out of space and looped back around to the top, so by 6:47 the notation flow is ambiguous.
you showed restraint in highlighting some failure modes of graphing calculators without dunking on them. I expect you know that you could graph look-alike replicas of your functions by offsetting them a tiny bit to shift the hole in the range onto a gap in the domain lattice that's used to sample and render the surface.
Thanks! I assume you mean "is it enough to check only smooth paths?" I think it depends on your coordinate system. If we're writing paths as y as a function of x then no, since x=0 isn't even a function. But in terms of polar coordinates, writing theta as a function of r, I'm pretty sure it is enough to check only smooth ones as long as you don't require smoothness at r=0. The function f(x)=e^(-1/(1-x^2)) is smooth on [-1,1] and all derivatives are 0 at -1 and 1, so if you have a sequence of points like at 7:02 you could piece together shifted and scaled copies of this to get a function theta(r) passing through the points.
@@josephnewtonf(x) is not defined at -1 or 1.
@@jessewolf7649 Yep, my mistake, that function is part of a piecewise function with f(x)=0 for |x|≥1
The king is back ! All hail the king !
4:45 caught me absolutly off guard and i've started laughing out loud and every body else can`t understand why
You didn’t specify it at the very end, but checking every curved path only works if you keep yourself to rational functions. Otherwise you can keep playing piecewise games.
Unless I’m mistaken, checking every curved path should work for piecewise functions as well. For the limit to not equal some value L, there must be some value E such that there is a point within every radius D of (0,0) for which the function differs from L by at least E. Taking D to be 1/2, 1/3, 1/4 and so on gives an infinite sequence of points, and then you can draw a spiralling path through them. So if the limit doesn’t equal L then it must not equal L along some path, and conversely if the limit is equal on every path then the overall limit must be the same.
@@josephnewton The simplest counterexample to the general statement: let f(x, y) = 1 at (0, 0) and 0 elsewhere. Any line terminating at (0, 0) also works.
For curves that pass through the origin, the only places where rational functions are nondifferentiable are at poles. So either your rational function is locally linear at the origin or it's a pole - either one would not fit general continuous piecewise functions. So I'm not sure that even that is a strong enough guarantee. (Not totally clear on this second paragraph though)
There's a difference between the limit at 0 existing and the function being continuous at 0; the piecewise example you provided is discontinuous, but the limit does exist (and equals 0).
@@wsshambaugh To add a bit to the other response, if you are looking at the limit as x aproaches a of f(x), the value f(a) doesn't matter(of course I implicitly assumed f is defined at a, but that doesn't matter either).
@@wsshambaugh Right, I forgot to include that the point within radius D can't be at (0,0). The full definition is that the limit at (x',y') equals L if for every E>0 there exists D>0 such that 0
Every upload for this channel demands time out of my day, as soon as I see the notification!
very nice video thats a funny function
I think for such a limit you would first try to identify how it curves (with some derivatives) and THEN try to do the limit
will you apply this video to SOME ?
That gap at zero; would most graphing calculators just kind of, fill it in with an approximation?
Even if it can't calculate 0, that gap is huge. It's like it's ignoring anything below 1. If it's just 0 it ignores, then it should just be a really tiny point.
The calculator only samples a grid of points, which are the vertices of the green checkerboard (that’s why there’s exactly 4 squares missing in the middle). I could have made the gap smaller by increasing the number of vertices, or just got rid of it by having (0,0) be between the sampled points, but the later functions in the video are a lot harder to graph accurately because of the big height changes
@@josephnewton ah I see, thank you!
Btw your “unnatural” counterexample also swiftly shows that existence of partial derivatives (they’re all 0 at the origin) does not imply differentiability (function is not even continuous at the origin, let alone differentiable).
A comment regarding the “check all curved paths” thing - one may show that the limit of f at, say, the origin exists, iff the limits of a FINITE family of restrictions of f to a collection of DISJOINT subsets whose union is the domain of f (all having the origin as an accumulation point blah blah) all exist and coincide. This is blatantly false for infinite families of disjoint subsets, e.g. the collection of straight lines through the origin.
However one may certainly consider an infinite family of non-overlapping CURVED paths that accumulate at the origin, along which the limit exists, and all of the limits coincide, but the overall limit still doesn’t exist. Take the unnatural counterexample again, and probe it along, say, curves of the form y = Ax^3 for real A, or logarithmic spirals that converge to 0…
Point is, the issue isn’t straight vs curved, but disjoint vs overlapping! Indeed, the theorem becomes true again if you allow the subsets to overlap.
What kinds of curved paths do you have to check? e.g. quartics? Can you just take a rational approximation or will that not work? For example ln(x)
Is it sufficient to check every rational path?
Well, I guess the same construction used on any non rational function will do!
@@tsawy6Maybe not, depends on how well we can approximate general functions using rational ones (kind of reminds me of Padè approximation)
@@Rotem_S wouldn't the same counter example he first presented with a quadratic for straight lines work just fine if we sub out the quadratic for e.g. an exponential?
The King 👑 returns LET HIM COOK 🔥🔥🔥
Isn't saying that "limit do not exist" just sweeping problem under the rug...?
Absolutely immaculate humor, well done
Golden video, now just invert the colors lol
I was taught that a limit of infinity is not the same thing as does not exist. You only get DNE if one side is positive infinity and the other side is negative infinity.
So, for example, the lim_x->0 (1/|x|) = infinity, but lim_x->0 (1/|x|) Does Not Exist.
i think you used |x| twice by mistake and meant lim x->0 1/x, and yes i agree
Great!
Why is half the video cut off?
Great video! There is a good problem in baby rudin that goes over this in chapter 4 I believe.
Our math prof actually preferred the definition of the limit agreeing on all paths over the epsilon delta definition.
Maybe I actually should have gone for math instead of compsi but oh well
yeah, I'm learning myself machine learning and most of the time when I get stuck it's because of not enough maths background
Loved the vid! Please keep uploading
3:00 bruh
now that I think of it I presume it's because x = 0 cannot really be represented as the form y = mx
yes x=0 epic
what if you actually check all lines though 3:47
4:20 oh
5:13 hump
I'm expecting x = i√a to go insane
6:47 bro....
i wish i watched this before taking calc 1 and 2
hey dude, i just stumbled on this video while walking the number line outside. are you a kiwi? awesome.
Funny enough this is the kind of thing that Ted kaczynski wrote about in his thesis. The points you’re describing are called “ambiguity points” and there can only be countable many of them regardless of the function
Can a function, at a point where it *is* defined, have a limit not equal to the function at that point? Or is the limit of a function at every defined point by definition the value of the function at that point?
Sure! That's just a non-continuous function, like f(x)={(1 if x=0) and (0 otherwise)}. The limit at 0 is 0, but the value at 0 is defined as 1!
Hmmm. I have no idea how this works
You know it's impressive you've made such a good video, I as a highschool student even understand it. Great job man.
Now you are guilty of not transforming (x, y) into (r, theta) when saying "Actually all" the lines 🔍🔎
Loving these videos, keep it up, even if it's a little late for me to pay more attention in my Real Analysis module
Is this real!? I never thought I’d see another video but I subbed last week just in case! Woo!
The epsilon delta definition is exactly the point where I stopped paying attention in the calculus class
#SoME3
"Oh, a function. Oh, a function. Oh, a function..."
I'm confused, why can't that piecewise polynomial function have a limit of 0? It's been a couple years since calculus, but I don't seem to remember "if the function is defined at x, the limit is either equal to f(x) or does not exist" being a rule.
Wouldn't f(x) = {0 if x = 0, 1 otherwise} have a limit of 0 as x goes to 0?
That’s correct, the limit can be different from the value of the function. But for the function f(x,y)=1 when x=y^2, there are points (x,y) arbitrarily close to (0,0) such that f(x,y)=1, so the limit can’t be zero
@@josephnewton Ah
This is why I found Calc 4 so cool :)
brilliant underrated video, keep making more!!
what is that real-time 3d graph builder that you are using?
i'm pretty sure it's the 'grapher' app that comes with mac computers; at least, mine produces graphs that look the same
3:30 where tf is x?
0:33 I keep hearing that this is a tricky thing to prove, but doesn't the Taylor series expansion of sin(x) just give it for free?
Or you could use L'Hopital's rule and take the derivative of the numerator and denominator to get cos(x)/1
Yes, what I meant was that I didn’t want to get into the technical details of how limits are actually defined, which is the epsilon-delta definition
depends on how you do it, but often the proof of sinx/x -> 1 will be a *prerequisite* for finding d(sinx)/dx, which you need for both the Taylor series and L'Hôpital's rule
Amazing!
loved this video
loved this video!
Nice!
Great video :).
I don't understand why the limit along the straight lines equals 0. You can get arbitrarily close to 1 along any of the straight lines, so it should equal 1.
You can get arbitrarily close in _distance_ to the point where the value is 1, but not in value. On all of those increasingly closer points, the value is still 0 (except where you actually cross the parabola) and does not approach 1. Instead, it jumps to 1 abruptly when you reach the end point. (Assuming that you were talking about the function at 4:08.)
@@jelmerterburg3588 so at that exact point the gradient is undefined and the function is discontinuous?
@@bscutajar Exactly :). The situation here is a bit more complicated than usual because there are two input variables and therefore infinitely many ways to move around the input plane. If you follow the function along the path of the parabola, or along a path which never crosses the parabola, the function is continuous for all points along that path. But if you consider the entire input plane (and all possible paths through it), the function is discontinuous.
About the sin(x)/x formalizing being messy.
No its not, its trivialy easy.
You start by noticing that sin(x)=x, so you get x/x reducing to 1.
I think x^y is a good example.
y = mx gives (x^x)^m which approaches 1.
x=0 gives 0^y, which gives 0.
and x = A^(1/y) gives A.
so most lines give you 1, one line gives you 0 and some curves give you any number you want.
My question is what set of functions give you the same limit when you approach 0 with a curve as with a line tangent to the curve. Note that different curves are still allowed to give different limits.
2:00 Actually when using interchanged implication, it is not sufficient just negating the sides, they need to be negated and interchanged at once, either interchanged implication method of proof is not valid