Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 :  f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\frac{1}{2}$ so the limit does not exist.
b)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{e^{xy^3}-1}{x^2 +y^4}$$
I don't have any ideas here
c)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = (x^2 +y^2)^{xy}$$
Let $x \geq y$ then $|x^2 +y^2|^{xy} \leq |2y^2|^{y^2} \longrightarrow 1 \quad $for $y \longrightarrow 0$ but for $x = \frac{1}{y} \Longrightarrow f(x,y) = (\frac{1}{y^2} + y^2)^1 \longrightarrow \infty \quad$ for $y \longrightarrow 0 $ so the limit does not exist. 

For b):
we have that $f(0,y) = f(x, 0) = 0$ for every $x,y\neq 0$.
If $x$ and $y$ are both non zero, we have that
$$
\frac{e^{xy^3} - 1}{x^2+y^4} = \frac{e^{xy^3} - 1}{xy^3}\cdot \frac{xy^3}{x^2+y^4}.
$$
The first factor at the r.h.s. is bounded (for $x,y$ bounded), since $(e^t-1)/t \to 1$ as $t\to 0$.
The second factor can be estimated by
$$
\left|\frac{xy^3}{x^2+y^4}\right| =
|y| \cdot \frac{|x| y^2}{x^2+y^4}\leq \frac{|y|}{2}.
$$
Hence the limit exists and is equal to $0$.
