Does the following limit exist? (a)$ \lim_{(x,y) \to (0,0)} f(x,y) = \frac{x^2+y}{x^2+y^2}$? Does the following limit exist? If exist, evaluate (a)$ \lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}$ $(b)\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}$?
for (a) 
approaching (0,0) using $y=0$ we have
$$\lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}\implies\lim_{x
\to 0} \frac{x^2}{x^2}= 1$$
approaching (0,0) using $y=-x^2$ we have 
$$\lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}\implies\lim_{x
\to 0} \frac{x^2 -x^2}{x^2+x^4}= 0$$
Since the limits are different, $\lim_{(x,y)
\to (0,0)} \frac{x^2+y}{x^2+y^2}$ $D.N.E$
for (b) approaching using $y=0$
$$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}\implies\lim_{x \to 0} \frac{x(0)}{x^2+0} =0 $$
approaching (0,0) from $x=y^2$
$$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}\implies\lim_{y \to 0}  \frac{y^4}{2y^4}=1/2 $$
Again limits are not equal so $\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2+y^4}$ $D.N.E$
Is this correct ?

we use $$x=\frac{1}{n},y=\frac{\sqrt{n}}{n}$$ and plug in this in your function
$$f(x,y)=\frac{x^2+y}{x^2+y^2}$$ we get
$${\frac {{n}^{2}+\sqrt {n}}{\sqrt {n} \left( n+1 \right) }}$$
and we obtain $$\lim_{n\to \infty}{\frac {{n}^{2}+\sqrt {n}}{\sqrt {n} \left( n+1 \right) }}=\infty$$ therefore the limit doesn't exist
