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When talking about limits for functions of several variables, why isn’t it sufficient to say, $$\lim_{(x,y)\to(0,0)} f(x,y)=L$$ if $f(x,y)$ gets close to $L$ as we approach $(0,0)$ along the $x$-axis ($y = 0$) and along the $y$-axis ($x = 0$)?

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    $\begingroup$ Because you have other possible paths than these two? $\endgroup$ – Clement C. Aug 5 '17 at 18:29
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    $\begingroup$ There are infinite paths approaching a point in $\mathbb R^2$. You're only considering two of them. Why the bias? $\endgroup$ – Sahiba Arora Aug 5 '17 at 18:29
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    $\begingroup$ @clementc. The notion of "a path" isn't embedded in the definition of a limit. But, certainly the limit fails to exist if along distinct paths the function approaches distinct values. $\endgroup$ – Mark Viola Aug 5 '17 at 18:32
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    $\begingroup$ @MarkViola I know. I merely pointed out that the suggested criterion failed to take into account the "same value on each path" idea that is in the very title. $\endgroup$ – Clement C. Aug 5 '17 at 18:33
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    $\begingroup$ @clementc. I thought that you knew given the high quality and depth of your posts. I just wanted to leave a comment that clarifies things a bit. $\endgroup$ – Mark Viola Aug 5 '17 at 20:22
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By definition, to show that $\lim_{(x,y) \to (0,0)}f(x,y) = L$, you have to show that for every sequence $(x_n, y_n) \to (0,0)$, you have $f(x_n,y_n) \to L$. Showing it for particular cases is not sufficient.

For example, let us look at $f(x,y)=\dfrac{xy}{x^2+y^2}$. For every $y \neq 0$ you have $f(0,y)=0$ and for every $x \neq 0$ you have $f(x,0)=0$. You want to conclude that $f \to 0$.

However for every $ n>0$, $f(1/n,1/n)=1/2$.

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    $\begingroup$ @boboquack because you want a sequence $\{(x_n,y_n)\}$ that approaches $(0,0)$. $\endgroup$ – AccidentalFourierTransform Aug 6 '17 at 9:22
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Consider $f(x,y)=\begin{cases}\frac{2xy}{x^2+y^2}, ~x^2+y^2\neq0\\0, ~x^2+y^2=0\end{cases}$.

This function is continuous along paths $x\to0,y=c$ and $x=c,y\to0$, but it is not continuous simultaneously since along the path $x=y$ its value is $1$, but in the origin it is defined as $f(0,0)=0$.

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    $\begingroup$ and to see what is really happening here, write $x=r \cos(\theta), y=r\sin (\theta)$ in which case the function takes the value $\sin (2\theta)$ when $r\not = 0$ $\endgroup$ – Henry Aug 6 '17 at 14:30
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If you mean every straight line path, the standard example without a limit is $$ \frac{x^2 y}{x^4 + y^2} $$ at points other than the origin. Limit at the origin is $0$ along the $y$ axis or along any line $y=mx$ through the origin. $$ \frac{m x^3}{x^4 + m^2 x^2} = \left( \frac{m}{m^2 + x^2} \right) \; \; x $$ where $$ \left| \frac{m}{m^2 + x^2} \right| \leq \left|\frac{1}{m} \right|$$ so that the absolute value of the function is no larger than $ \left|\frac{x}{m} \right|$ along that line.

HOWEVER: One value when $y = x^2,$ different when $y = -x^2$

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  • $\begingroup$ (+1) I was going to post an example of a limit that exists and equals the same value for each linear path, but not on some other path. $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 21:36
  • $\begingroup$ @SimplyBeautifulArt thanks. I think it likely that, given a bound $n,$ there is an example giving limit zero along all paths $(x,y) $ through the origin with $x(t)$ and $y(t)$ both polynomials of degree no larger than $n.$ I am trying to remember whether we can expand to, say, real analytic paths through the origin. $\endgroup$ – Will Jagy Aug 5 '17 at 21:48
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    $\begingroup$ Oh jeez, that sounds scary :) $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 21:59
  • $\begingroup$ @SimplyBeautifulArt scary? Perhaps - though I would love to see a proof of this claim! $\endgroup$ – Brevan Ellefsen Aug 6 '17 at 7:07
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    $\begingroup$ @WillJagy How about $\tfrac{y \exp(-1/x^2)}{y^2+\exp(-2/x^2)}$? If $y(t)$ is identically $0$, then the limit is $0$. If not, and $x(t)$ and $y(t)$ are both real analytic functions approaching $0$, then $\exp(-1/x(t)^2)/y(t) \to 0$ so the ratio goes to $0$. However, putting $y = \exp(-1/x^2)$ gives $1/2$. $\endgroup$ – David E Speyer Aug 6 '17 at 12:35
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If you go back to the definition (for every eps > 0, there is a delta > 0 etc. etc. ) then it seems a priori quite possible that for every path the limit exists and is the same, but for some eps > 0 there is no delta > 0 that works for all paths simultaneously.

It all depends on how you define "path". If you only consider two paths "along the x-axis" and "along the y-axis", or only any straight line paths, I'd be confident that there are counter examples.

For example, take all paths that are straight lines towards (0, 0). Each approaches (0, 0) at an angle alpha with 0 < alpha ≤ 360 degree. Calculate f (x, y) by calculating the angle (0 < angle ≤ 360 degree) and the distance d, and let f (x, y) = d / angle. On every straight towards (0, 0) the limit is 0, but in any neighbourhood of (0, 0) there are arbitrarily large values of the function.

So to prove that the limit exists, you will need to allow more paths than just straight lines. My counter-example wouldn't work if you allowed paths that approach (0, 0) in a spiral that goes around (0, 0) infinitely often - along such a path the limit wouldn't exist. (For example takes the paths where (x, y) = (sin (t + d) / t, cos (t + d) / t) for different values of d, and t going from 1 to infinity. Again a counter example could be produced by rotating the original counter example function to "unbend" all these paths into straight lines).

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You are basically asking why: $$\lim_{(x,y)\to(0,0)} f(x,y)=L\ne \lim_{x\to 0} \lim_{y=0} f(x,y) \ \ \ ?$$ Because the inside limit may/may not be defined and may/may not exist.

Consider the function: $f(x,y)=(1+x)^{\frac1y}$. We want to find: $$\lim_{(x,y)\to(0,0)} (1+x)^{\frac1y}.$$ Note: $$\lim_{y=0,x\to 0} (1+x)^{\frac1y}=undefined,$$ $$\lim_{x=0,y\to 0} (1+x)^{\frac1y}=1,$$ $$\lim_{x=y \to 0} (1+x)^{\frac1y}=e,$$ $$\lim_{\left(x=\frac{1}{n},y=\frac{1}{n^2},n\to\infty\right)} (1+x)^{\frac1y}=\infty,$$ $$\lim_{\left(x=\frac{1}{n^2},y=\frac{1}{n},n\to\infty\right)} (1+x)^{\frac1y}=1.$$ You can consider many other sequences (paths) to see the different results.

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$\mathbb{R}^2 ,\forall n \in \mathbb{N}$ is a metric space with the metric $d(x,y)=\sqrt{(x_1-y_1)^2+....+(x_n-y_n)^2}$

Every metric space $(X,d)$ has the Hausdorff seperation property i.e:

$\forall x,y \in X,x \neq y$ we can find open balls $B(x,\delta_1),B(y,\delta_2)$ such that $$B(x,\delta_1) \cap B(y,\delta_2)= \emptyset$$

Now with this propert you can prove as an exercise that a limit of a function $f:(X,d_1) \rightarrow (Y,d_2)$ or the limit of a sequence $x_n \in X$ is unique.

Thus for a function in $\mathbb{R}^n$ its limit if it exists,is independent of the path we use to approximate the value of the function at a given point,because it must be unique.

You can keep this answer in mind if you have encountered the definition and some properties of metric spaces.

If you don't then you can ignore it and just prove that a function from $\mathbb{R}^n$(in this case $\mathbb{R}^2$) to $\mathbb{R}$ cannot have two different limits at a given point.

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  • $\begingroup$ You can produce italicized text simply by wrapping it with asterisks *. If you wrap it with dollar signs, it is treated as a math formula. Implying, among other things, that individual letters are assumed to stand for distinct variables. Therefore the spacing between the letters is out of whack. $\endgroup$ – Jyrki Lahtonen Aug 8 '17 at 10:38
  • $\begingroup$ ok thank you..i will remember that. $\endgroup$ – Marios Gretsas Aug 8 '17 at 10:47

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