# How to prove the limit exists for function of two variables?

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.

• The first reasoning is correct, well done. The second (Squeeze) has an easily fixed flaw at points $(0,y)$. – André Nicolas Jul 28 '15 at 12:23
• alternativly you may use polar coordiantes – tired Jul 28 '15 at 12:32
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 $0$.