Find the limit to $\lim\limits_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$ My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the expression. My "toolbox" for solving problems like these is very limited... If polar form doesn't work, then I usually have no clue on how to continue.

Edit: I think this is the solution. 
$$
\lim_{(x,y)\to(0,0)}\left|\frac{x^5+y^2}{x^4+|y|}\right| = \frac{|x^5+y^2|}{|x^4+|y||}
$$
Applying the triangle inequality gives
$$
\frac{|x^5+y^2|}{|x^4+|y||} \leq \left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right|
$$
Inspecting the denominators on the RHS gives:
$$
\left|\frac{x^5}{x^4+|y|}\right| \leq |x|, \quad\left|\frac{y^2}{x^4+|y|}\right| \leq |y|
$$
So
$$
\left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right| \leq |x| + |y|
$$
Since $|x| + |y| \to 0$ when $x,y\to 0$, the sandwich theorem states that $|\frac{x^5+y^2}{x^4+|y|}| \to 0$. And if $\lim |f(x)|=0$ then $\lim f(x)=0$ which solves the original problem.
Hint:)

For paths $x=0$ as $y\to0$ and $y=0$ as $x\to0$ the result of limit clearly is zero, furthermore
$$\Big|\frac{x^5+y^2}{x^4+|y|}\Big|\leq\Big|\frac{x^5}{x^4+|y|}\Big|+\Big|\frac{y^2}{x^4+|y|}\Big|\leq|x|+|y|$$
A common tool is to homogenise the denominator by setting $\displaystyle u=\frac{y}{x^4}$, then discuss the limit in function of $u$.

$\displaystyle f(x,y)=\frac{x^5+y^2}{x^4+|y|}=\frac{x^5+u^2x^8}{x^4+|u|x^4}=\frac{x+u^2x^4}{1+|u|}=\frac{x+uy}{1+|u|}=x\left(\frac 1{1+|u|}\right)+y\left(\frac u{1+|u|}\right)$

Since both $\frac 1{1+|u|}$ and $\frac u{1+|u|}$ are bounded (by $1$ in absolute value)

Then $|f(x,y)|\le |x|+|y|\to 0$