I get how for the limit as (x,y) approaches, say, (0,0), if there's a discontinuity along, say, x=0, then there won't be a limit, because no matter how small you make your $\delta$-circle, you won't get within $\epsilon$ of a limit. But I don't get how f could approach the same limit along every line through the point in question, but not approach the limit along another path. Could someone give an intuitive explanation (mathematical explanations are good too, but intuitive ones are preferred) of what a "path" means in terms of limits and how a limit could be the same within any $\delta$ along every line, but how the limit doesn't exist because it is different along a different non-linear path? If possible please include some mention of the $\delta$-circle.
3 Answers
Here is an example which, I hope, is graphical enough. Consider 2 circles: the circle of radius 2 with the center at (2,0) and the circle of radius 1 with the center at (1,0). These circles touch at (0,0), with the smaller circle lying inside the bigger one.
Now consider a function which is equal to 0 outside of the larger circle, equal to 0 inside the smaller circle, and equal to 1 between the two circles.
If you approach (0,0) along the vertical line, then you're outside of the larger circle, so the value of the function is 0. If you approach (0,0) along a non-vertical straight line "from the left" (i.e., from the half-plane $x<0$), then you're outside of the larger circle, so the value of the function is 0. If you approach (0,0) along a non-vertical straight line "from the right", then at some point you enter the inner circle, and after that the value of the function is 0. So if you approach (0,0) along any straight line, the limit of the function is 0.
However, you can also approach (0,0) along a curve which lies between the two circles, for example, along the circle of radius 1.5 with the center at (1.5, 0). And then you get the limit 1, as the function is equal to 1 between the circles.
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This is a great example, why is it not awarded already? ha : ) $~~$ Having said that, it can be perfected if someone is so kind as to furnish it with a figure or even animation. – Lee David Chung Lin Sep 28 '16 at 07:01
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Let's work in polar coordinates. Take $\theta$ to be the angle. Now suppose the value of a function at $(r,\theta)$ is $r/\theta$ for $\theta$ non-zero and $0$ when it is. If we go down a ray we definitely converge to zero.
However, if on a path $\theta$ goes to zero faster than $r$ does, we get infinity.
eg
define $x_n = (n^{-1},n^{-2})$ then the function has $f(x_n) = n.$
Here we are on a path lying in a set $\theta =r ^2.$
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That makes sense algebraically, and I've seen other similar examples that make algebraically... but it still doesn't make sense to me intuitively/graphically - why doesn't an infinite amount of lines fill the circle around (0, 0) with a radius of δ (see comments for an elaboration)? – Adam Zerner Sep 25 '16 at 00:04
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Here's a thing to think about for the given function. Yes, for every line, the limit heading into the origin is zero. However, each of those lines (for $\theta \ne 0$) has a point on in for which the function is equal to 1. The function evaluated on this path always has a value of 1 (again for $\theta \ne 0$), so the limit if you go along it with descending $\theta$ is 1. – Davis Yoshida Sep 25 '16 at 01:09
Let $f:B \rightarrow \mathbb{R}$, where $B$ is an open ball in $\mathbb{R}^2$ with center $a$.
By definition, $\lim_{x \rightarrow a} f(x) = L$ means that for every $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$.
However, there is an equivalent criteria for $\lim_{x \rightarrow a} f(x) = L$ in terms of sequences. You can find it in basically any undergraduate analysis textbook (like Rudin, Apostol, Bartle, Abbot, Pugh).
$\lim_{x \rightarrow a} f(x) = L$ if and only if $\lim_{n \rightarrow \infty} f(x_n) = L$ for every sequence $(x_n)$ with $\lim_{n \rightarrow \infty} x_n = a$ and $x_n \neq a$.
The elements $x_1,x_2,x_3,\ldots,$ of a sequence satisfying $\lim_{n \rightarrow \infty} x_n = a$ and $x_n \neq a$ trace out a "path" towards $a$. Notice that the path does not need to be linear. Imagine a path that spirals in towards $a$ for example.
If you want to show the limit of $f$ doesn't exist, and you want to be very rigorous about it, you would find a sequence $(x_n)$ satisfying $\lim_{n \rightarrow \infty} x_n = a$ and $x_n \neq a$ but $\lim_{n \rightarrow \infty} f(x_n) \neq L$ (say because $f(x_n) = L+1$ for every $x_n$ in your sequence).
This limit criteria in terms of sequences is a bit complicated for most calculus classes. So we just talk about the paths instead.
The comments of Adam Zerner seem to be about whether an arbitrarily small ball around the point $a$ can be filled up by the set of all line segments that start at the edge of the ball and end at $a$. The ball indeed can be filled up in that way.
But the limit is not about filling up a ball. It is about what happens as you move closer and closer towards $a$. For the limit to exist, for every distance $\epsilon > 0$, you must be able to find a ball of radius $\delta$ around $a$ so that no matter which point $x$ you look at in the $\delta$-ball, you have the distance between $f(x)$ and $L$ less than $\epsilon$. You can look at $x$ to the left or to the right of $a$ or above or below $a$ or on some path or line segment towards $a$, just as long as $x$ is within radius $\delta$ of $a$.
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Socrates gives us an ϵ and says, "give me a circle in the x-y coordinate plane (input space) such that any point in that circle will give me a value within ϵ of the limit you're claiming to exist".
So ultimately, the question is, can we draw a bunch of lines such that it'd completely fill the circle?
Try drawing a lot of lines. Then zoom in, and there'll be space. Then fill that space with more lines. Then zoom in more.
"You could always draw more lines."
"But you could always zoom in more"
...
– Adam Zerner Sep 24 '16 at 23:41