$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $ Does it exist or not? I have this limit: 
$$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $$
I think that this limit does not exist (and wolfram|alpha agrees with me). But I can't find a way to prove it. I chose some paths and the limit was always equal to zero. I tried polar coordinates but I couldn't get an answer. Any tips?
(I also have this one: $$ \lim_{(x,y)\rightarrow (0,0)} x^{y^{2}} $$ 
Can I get 2 different paths to show that the limit does not exist? 
$$x=0: \lim_{(x,y)\rightarrow (0,0)} 0^{y^{2}} = \lim_{(x,y)\rightarrow (0,0)} 0 = 0,  $$ $$y=0: \lim_{(x,y)\rightarrow (0,0)} x^{0^{2}} = \lim_{(x,y)\rightarrow (0,0)} 1= 1. $$ Am I right? )
 A: The first limit actually exists and is zero. You can show that by definition.
Namely,
$$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 = \frac{|x^5 y^3|}{x^6+y^4} \cdot \frac{|x^5y^3|}{x^6+y^4}$$
Because
$$x^6+y^4\ge x^6 \implies \frac{1}{x^6+y^4} \le \frac{1}{x^6}$$
$$x^6+y^4\ge y^4 \implies \frac{1}{x^6+y^4} \le \frac{1}{y^4}$$
assuming $x$ and $y$ are both non-zero, we have that
$$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 \le \frac{|x^5 y^3|}{x^6} \cdot \frac{|x^5y^3|}{y^4} = x^4y^2$$
Taking the square root of both sides we get
$$\left|\frac{x^5y^3}{x^6+y^4}\right| \le x^2|y|$$
You can see that this inequality also holds when one of $x$, $y$ is zero (both cannot be zero at the same time). For this to be smaller than $\epsilon$ it suffices to take $\delta=\sqrt[3] \epsilon$.
The second limit does not exist and your proof is correct.
A: a quick trick worth trying :
$$ (x^3 - y^2)^2 \geq 0, $$
$$  x^6 + y^4 \geq 2 x^3 y^2 . $$
$$ (x^3 + y^2)^2 \geq 0, $$
$$  x^6 + y^4 \geq -2 x^3 y^2 . $$
$$ \color{purple}{  x^6 + y^4 \geq 2 |x|^3 y^2 } . $$
when $x,y$ are not both zero,
$$ \frac{1}{2}  \geq  \frac{ |x|^3 y^2}{x^6 + y^4} . $$
$$ \frac{x^2 |y|}{2}  \geq  \frac{ |x^5 y^3|}{x^6 + y^4} . $$
