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Problem: Evaluate the indicated limit or explain why it does not exist: \begin{align*} \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2 + y^4} \end{align*}

The definition of limit my calculus textbook gives is:

We say that $\lim_{(x,y) \to (a,b)} f(x,y) = L$, provided that:

1) Every neighbourhood of $(a,b)$ contains points of the domain of $f$ different from $(a,b)$, and

2) For every positive number $\epsilon$ there exists a positive number $\delta = \delta (\epsilon)$ such that $|f(x,y) - L| < \epsilon$ holds whenever $(x,y)$ is in the domain of $f$ and satisfies $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$.

So in this specific problem I first checked the limiting behaviour when $(x,y)$ approaches $(0,0)$ from different directions:

When $y =0$ or $x = 0$ then $\lim_{(x,y) \to (0,0)} f(x,y) = 0$. When $y = x^2$, we also have that $\lim_{(x,y) \to (0,0)} f(x,x^2) = 0$.

So I believe the limit exists and is zero, and I want to prove it now by using the definition. So let $\epsilon > 0$. Then we need to find a $\delta > 0$ such that if $0 < \sqrt{x^2 + y^2} < \delta$, then $| \frac{x^2 y^2}{x^2 + y^4} - 0 | < \epsilon$.

Now, since $x^2 \leq x^2 + y^4$, it follows that $\frac{x^2}{x^2 + y^4} \leq 1$. Let $\delta = \sqrt{\epsilon}$. If $0 \leq \sqrt{x^2 + y^2} < \delta$, then it follows that $x^2 + y^2 < \epsilon$. Thus we also have that $y^2 < \epsilon$.

Since we already had that $\frac{x^2}{x^2 + y^4} \leq 1$ we can multiply this with $y^2$ and get $\frac{x^2 y^2}{x^2 + y^4} \leq y^2$, from which it follows that $\frac{x^2 y^2}{x^2 + y^4} < \epsilon$.

Can someone tell me if my reasoning is correct? Also, is this the right method to proof the existence of limits of functions of two variables? I mean, if you suspect that the limit exists, you have to use the delta-epsilon notation to prove it?

Also, I found an alternative solution:

Since $0 \leq \frac{x^2 y^2}{x^2 + y^4} \leq \frac{x^2 y^2}{x^2}$, and since $\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2} = 0$, we have also $\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2 + y^4} = 0$ by the Squeeze Theorem.

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  • $\begingroup$ The first reasoning is correct, well done. The second (Squeeze) has an easily fixed flaw at points $(0,y)$. $\endgroup$ Jul 28, 2015 at 12:23
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    $\begingroup$ alternativly you may use polar coordiantes $\endgroup$
    – tired
    Jul 28, 2015 at 12:32
  • $\begingroup$ If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. $\endgroup$
    – gebruiker
    Apr 28, 2016 at 19:43

1 Answer 1

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Let's to use polar coordinates $x=\rho\cos\phi$, $y=\rho\sin\phi$, and your limit is $$ \lim_{\rho\to0} \frac{\rho^4 \cos^2\phi \sin^2\phi}{\rho^2\cos^2\phi + \rho^4\sin^4\phi} = \lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi} $$ But $\lim_{\rho\to0}\rho^2 \cos^2\phi \sin^2\phi = 0$. If $\cos\phi\ne0$, then $$ \lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi} = \frac{\lim_{\rho\to0}(\rho^2 \cos^2\phi \sin^2\phi)}{\lim_{\rho\to0}(\cos^2\phi + \rho^2\sin^4\phi)} = \frac{0}{\cos^2\phi} = 0. $$ If $\cos\phi=0$, $$ \lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi} = \lim_{\rho\to0} \frac{0}{\rho^2} = 0 $$ In any case, limit equal to $0$.

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