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I've been asked to show that the limit $$\lim \limits_{(x,y) \to (0,0)}\frac{x^2y^2}{2x^2+y^2}$$ exists. So far I have managed to deduce that the function will $\to 0$ as $(x,y)\to0$. This is because letting $y=0, x=0, y=x,$ etc will yield $0$ as the limit. However, I'm still having problems proving the limit exists as the fore mentioned method really isn't proof..

If someone has an idea how to prove this say using the epsilon-delta-method, any help would be greatly appreciated. Thanks!

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    set $y=r \sin(\phi),x=\frac{r}{\sqrt{2}}\cos(\phi)$ – 2017-01-10

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Write it in polar coordinates by substituting for $x=rcos(t)$ and $y=rsin(t)$. The bottom will then be simply 2$r^2$ and you can eliminate the denominator with the $r^4$ of the numerator. The expression becomes:

$r^2$$(cos(t)sin(t))^2$/2

for which equals r^2(sin(2t))^2/8

x and y tends to zero is equivalent to r and t both tending to 0. By looking at that expression, you can deduce that the limit exists for all directions of $t$ and it always tends toward 0.

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    good but need fixing – 2017-01-10
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You have $2x^2 + y^2 \ge y^2$. Hence:

$$ \left | \frac{x^2y^2}{2x^2+y^2} \right | \le | x^2| = x^2$$

Now you can apply either Squeeze Theorem or $\epsilon - \delta$ definition of limit to finish the prood.

For the $\epsilon - \delta$ proof you can use that: $x^2 \le y^2 + x^2 = \left(\sqrt{x^2+y^2}\right)^2 < \delta^2$. So choosing $b=\sqrt{\epsilon}$ would do the trick.

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Try polar coordinates: $$0\le\frac{x^2y^2}{2x^2+y^2}=\frac{r^4\cos^2\theta\sin^2\theta}{2r^2\cos^2\theta+r^2\sin^2\theta}=\frac{r^2\sin^22\theta}{4(1+\cos^2\theta)}\le\frac{r^2}4\to0$$

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So far I have managed to deduce that the function will $\to 0$ as $(x,y)\to0$. This is because letting $y=0, x=0, y=x,$ etc will yield $0$ as the limit.

Approaching zero along those paths is not sufficient to conclude that $f(x,y) \to 0$ as $(x,y) \to 0$. The statement that $f(x,y) \to 0$ as $(x,y) \to 0$ is equivalent to $\lim_{(x,y)\to(0,0)} f(x,y) = 0$.

Here is one way to show the limit is zero. Since $$ 0 < x^2 < 2 x^2 < x^2 + y^2 < 2x^2 + y^2 $$ for all $(x,y)\neq (0,0)$, we have $$ 0 < \frac{x^2}{2x^2 + y^2} < 1 $$ Therefore $$ \left|\frac{x^2y^2}{2x^2 + y^2}\right| = \left|\frac{x^2}{2x^2 + y^2}\right|\left|y^2\right| < 1 \cdot |y^2| = y^2 $$ Since $y^2 \to 0$, the two-variable limit is zero.