# proving that the following limit exist

How can I prove that the following limit exist? $$\mathop {\lim }\limits_{x,y \to 0} \frac{{x^2 + y^4 }} {{\left| x \right| + 3\left| y \right|}}$$ I tried a lot of tricks. At least assuming that this limit exist, I can prove using some special path (for example y=x) that the limit is zero. But how can I prove the existence?

Hint:

Try using the squeeze theorem. Since everything is positive, you're already half way there, that is if the limit exists, we know $$0\le \lim_{(x,y)\to (0,0)}\frac{x^2+y^4}{|x|+3|y|}$$

For the other direction split the fraction into two parts, and try using the inequality:

$$\frac{x^2}{|x|+3|y|}\le|x|$$

A similar inequality will work for the $y^4$ part of the fraction.

We choose the norm one: $$||(x,y)||=|x|+|y|$$ and we have

$$|x|\leq||(x,y)||\quad;\quad |y|\leq||(x,y)||$$ so $$0\leq\frac{{x^2 + y^4 }} {{\left| x \right| + 3\left| y \right|}}\leq\frac{||(x,y)||^2+||(x,y)||^4}{||(x,y)||}=||(x,y)||+||(x,y)||^3$$

There are more appropriate ways, but let's use the common hammer. Let $x=r\cos\theta$ and $y=r\sin\theta$. Substitute. The only other fact needed is that $|\sin\theta|+|\cos\theta|$ is bounded below. An easy lower bound is $\frac{1}{\sqrt{2}}$.

When you substitute for $x$ and $y$ on top, you get an $r^2$ term, part of which cancels the $r$ at the bottom, and the other part of which kills the new top as $r\to 0$.