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Prove that the limit $$\lim\limits_{(x,y)\to(0,0)}\frac{2x^2 y}{ x^4 + y^2}$$ when doesn't exist with an $\varepsilon$-$\delta$ argument.

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Please tell us what you have already tried, and what difficulty you have experienced. –  user22805 Sep 26 '12 at 3:35
    
I have tried to find some \delta, but couldn't find –  Frank Xu Sep 26 '12 at 3:37
    
That is what you would do if you were trying to find the limit, or to prove that the limit exists. Here, you need to do the opposite - show that no matter what $\delta$ is, you can't force the expression fall within $\epsilon$ of some value by choosing $x$ and $y$ within $\delta$ of 0. –  user22805 Sep 26 '12 at 3:45
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2 Answers

up vote 3 down vote accepted

Upshot: The idea here is that you want to choose two different paths to the origin--which will allow you to rewrite in terms of only one variable--and use delta-epsilon to prove what the limits along those paths are. To show that the more general limit (that is, not restricted to a particular path) fails to exist, you need to pick your two paths so that the limits along those paths are not the same. The key result you'll need to know (and use) here is that a real-valued function cannot converge to distinct limits as we approach a given point.

A nice way to go here is to choose functions $y=f(x)$ and $y=g(x)$ for your two paths such that $f(0)=g(0)=0$, and such that you get some nice cancellation to allow easy limit evaluation (and consequently, easier delta-epsilon work). In particular, it'd be excellent if we could either get the exponents of $x$ and $y$ to match in the denominator or get rid of one of $x,y$ altogether. For example, if we move along the curve $y=\alpha x^2$ for some constant $\alpha$, then $(x,y)\to(0,0)$ if and only if $x\to 0$. Then substituting $y=\alpha x^2$, we have $$\lim_{(x,\alpha x^2)\to(0,0)}\frac{2x^2\cdot\alpha x^2}{x^4+(\alpha x^2)^2}=\lim_{x\to 0}\frac{2\alpha x^4}{(1+\alpha^2)x^4}=\lim_{x\to 0}\frac{2\alpha}{1+\alpha^2}=\frac{2\alpha}{1+\alpha^2}.$$ This works for any alpha, and hopefully you can see that different $\alpha$ can result in different limits. Thus, the general limit cannot exist. For an altogether different example, we could approach along the line $x=0$.

What I've shown above isn't the rigorous delta-epsilon proof you need, but the fact that all variables disappeared in the process of evaluating the limit should indicate just how trivial the delta-epsilon proof will actually be....

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So in order to prove with the $\delta$ method, is it OK to just repeat what you showed, and pick a value randomly, and set it to Limit, and prove that other values have the different limits, no matter what $\delta$ I choose? –  Frank Xu Sep 26 '12 at 4:38
    
It sounds like you've probably got it figured out. Let me put it another way to be sure: Find some $\alpha,\beta$ such that $\cfrac{2\alpha}{1+\alpha^2}\neq\cfrac{2\beta}{1+\beta^2}$, and set $\varepsilon$ less than half of $\left|\cfrac{2\alpha}{1+\alpha^2}-\cfrac{2\beta}{1+\beta^2}\right|$. Suppose by way of contradiction that the limit exists, say $L$. Show that for any $\delta>0$, there is some $x\neq 0$ such that both $(x,\alpha x^2)$, $(x,\beta x^2)$ are within $\delta$ of $(0,0)$. By triangle inequality, the function values are within $2\varepsilon$ of each other. Contradiction. –  Cameron Buie Sep 26 '12 at 14:54
    
For simplicity, you may as well take $\alpha=1$, $\beta=0$. –  Cameron Buie Sep 26 '12 at 16:43
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Hint: define $z=x^2$. Have you seen a problem with $\lim\limits_{(x,y)\to(0,0)}\frac {xy}{x^2+y^2}$?

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No. I haven't used delta-eplison to prove some limit doesn't exist. –  Frank Xu Sep 26 '12 at 3:40
    
@FrankXu: what happens if $x=y?$ –  Ross Millikan Sep 26 '12 at 3:42
    
How can I do this kind of problem? Thank you! –  Frank Xu Sep 26 '12 at 3:44
    
Yeah, I know how to do this problem by setting $x = y^2$, but how to do this with $\eplison$ and $\delta$? –  Frank Xu Sep 26 '12 at 3:45
    
@FrankXu: you are backwards in my hints. If you put $x^2=y$ into your original problem, what happens? –  Ross Millikan Sep 26 '12 at 3:47
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