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?
 A: 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.
A: We choose the norm one:
$$||(x,y)||=|x|+|y|$$
and we have
$$|x|\leq||(x,y)||\quad;\quad |y|\leq||(x,y)||$$
so
$$0\leq\frac{{x^2  + y^4 }}
{{\left| x \right| + 3\left| y \right|}}\leq\frac{||(x,y)||^2+||(x,y)||^4}{||(x,y)||}=||(x,y)||+||(x,y)||^3$$
A: 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 $r\to 0$. 
