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I am lost on this problem:

State whether the following limit exists and prove it:

$$ \lim_{(x,y) \rightarrow (0,0)} \frac{\sqrt[2]{|x|}y}{x^2+y^2} $$

All of the examples in class used the $\epsilon$ - $\delta$ proof technique. I am still getting used to this proof technique and I understand what $\epsilon$ and $\delta$ represent, but I don't know how to begin (or end) $\epsilon$ - $\delta$ proofs.

What I know: Trying to prove $\forall\epsilon>0, \exists \delta$ such that $|\sqrt[2]{x^2+y^2}|<\delta \implies |\frac{\sqrt[2]{|x|}y}{x^2+y^2}-?|<\epsilon$

Specific questions: How do I proceed with the proof if I don't know what ? is? What does $|f(\mathbf x)-\mathbf a|<\epsilon$ mean in a multivariable case? What are general $\epsilon$ - $\delta$ proof techniques?

I think if I can figure out this problem I can figure out the rest.

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  • $\begingroup$ For problems of this kind it helps to write $x=r\cos \phi$, $\>y=r\sin\phi$ and to see what can happen when $r\to0+$. $\endgroup$ – Christian Blatter Sep 10 '13 at 16:27
  • $\begingroup$ @NicholasR.Peterson You're right. I gotta pay more attention. $\endgroup$ – Git Gud Sep 10 '13 at 16:50
  • $\begingroup$ @GitGud You should see some of the comments/answers I've posted before my morning coffee has kicked in :-) $\endgroup$ – Nick Peterson Sep 10 '13 at 16:55
  • $\begingroup$ @NicholasR.Peterson $\ddot \smile$ $\endgroup$ – Git Gud Sep 10 '13 at 16:59
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With multivariable limits, the thing to remember is this: the limit can only exist if the path limits along all paths exist and are equal, in which case the multivariable limit in and the path limits agree.

So, to find your "?", start by taking a path limit. For instance, along the line $y=x$, we have $$ \lim_{x\rightarrow0}\frac{\sqrt{\lvert x\rvert}x}{2x^2}=\lim_{x\rightarrow0}\frac{\sqrt{\lvert x\rvert}}{2x} $$ But this limit doesn't exist! Approaching $(0,0)$ along $y=x$ for $x<0$ you get $-\infty$, and approaching $(0,0)$ along $y=x$ for $x>0$ you get $\infty$.

So, you need to switch it up here: you need to show that the limit does not exist!

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  • $\begingroup$ Thanks! As long as the limit does not exist, epsilon-delta proofs aren't necessary, right (assuming you can find a contradiction using path limits)? Since the limit does not exist for one path it cannot exist for any paths, correct (if it does then there is a contradiction on the limit being unique)? $\endgroup$ – user2154420 Sep 10 '13 at 22:33
  • $\begingroup$ That's the intuition. To be more formal, I've given you a blueprint for showing that given any $\epsilon>0$, for every $\delta>0$ there exists two points that have distance at most $\delta$ from each other and differ by more than $\epsilon$ -- showing that no limit can exist. $\endgroup$ – Nick Peterson Sep 10 '13 at 23:45

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