How do you find the limit of $\frac{4x^4 + 5y^4}{x^2 + y^2}$? Find the limit of
$\frac{4x^4 + 5y^4}{x^2 + y^2}$ as $(x,y)\to (0,0)$.
Which method do I use to find the limit of that? I tried paths but the limits all came out to be $0$... (as a side question, when do you stop trying paths? I mean there are so many ways to try out when $x$ approaches $0$. You can try $y=0$, $y=x$, $y=x^2$, $y=mx$, and so many more ways. After you get like $0$ for 4 limits, do you just stop there and assume to try another method?) (Also, when I try different ways for paths, will the limits always be either $0$ or a finite number and never DNE?) 
Thank you
 A: In principle, you can never stop looking for new paths and may have to be creative.
However, you are right to suspect that the limit is indeed $0$.
To show this, you better prove that for each $\epsilon>0$ there exists $\delta>0$ such that $\sqrt{x^2+y^2}<\delta$ implies $|f(x,y)|<\epsilon$.
For your second question: Virtually anything can happen along different paths - different limits, divergence to infinity, and and of course proper divergence (think of zigzagging between two paths with different limits - the zigzag path will not have a limit).
A: Is this what you mean?
$$\underset{x,y\rightarrow0}{\lim}\frac{4x^4+5y^4}{x^2+y^2}=$$
$$=\underset{x,y\rightarrow0}{\lim}\frac{4x^4}{x^2+y^2}+\underset{x,y\rightarrow0}{\lim}\frac{5y^4}{x^2+y^2}=$$
$$=\underset{x\rightarrow0}{\lim}\frac{4x^4}{2x^2}+\underset{y\rightarrow0}{\lim}\frac{5y^4}{2y^2}=$$
$$=\underset{x\rightarrow0}{\lim}2x^2+\underset{y\rightarrow0}{\lim}\frac{5}{2}y^2=0+0=0$$
A: The euclidean norm is
$$||(x,y)||=\sqrt{x^2+y^2}$$
and we have
$$x^4\leq||(x,y)||^4\quad;\quad y^4\leq||(x,y)||^4$$
so we have
$$0\leq\frac{4x^4 + 5y^4}{x^2 + y^2}\leq9||(x,y)||^2\to_{(x,y)\to(0,0)}0$$
A: There is a nearly universal strategy when the denominator is $x^2+y^2$ or a close relative. Let $x=r\cos \theta$, $y=r\sin\theta$. 
Here the bottom becomes $r^2$, and the top is $4r^4\cos^4\theta+5r^5\sin^5\theta$. Divide. We get  $4r^2\cos^4\theta+5r^3\sin^5\theta$. The trigonometric functions are bounded, so the limit is $0$.
A: Well , let $x^2+y^2=r^2$  then   $$\frac{4x^4 + 5y^4}{x^2 + y^2}\leq\frac {5 r^4}{r^2} =5r^2 \to 0$$  as  $r \to 0$   $\hspace{99mm} \blacksquare$
A: The standard procedure is indeed changing the coordinates from Cartesian to Polar.  Your limit:
$$
\lim_{(x, y)\to(0, 0)} \frac{4x^4 + 5y^4}{x^2 + y^2}
$$
Will become (with $x=r\cos(\theta)$ and $y=r\sin(\theta)$ and using the fact that $r\to0$ when $(x, y)\to(0, 0)$):
$$
\lim_{r\to0} \frac{4r^4\cos(\theta)^4 + 5r^4\sin(0)^4}{r^2\cos(\theta)^2 + r^2\sin(\theta)^2}
$$
which ONLY exists if it is the same for all values of $\theta$ (the limit does not vary with $\theta$).
Using: $\sin^2(\theta)+\cos^2(\theta)=1$ your limit will further reduce to:
$$\lim_{r\to0} 4r^2 \to 0 \text{ for all } \theta \text{; therefore the limit exists.}$$
