Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$ 
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$

If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!

Recall Taylor series : $$\sin^2 y =y^2+O(y^3) ,\ \ln\
(1+x^2)=x^2+O(x^3)
$$ so that there is $\delta_i$ s.t. $$ 1-\epsilon < \frac{y^2}{\sin^2y},\ \frac{x^2}{\ln\ (1+x^2)}<1+\epsilon $$ for $|y|<\delta_1$ and $|x|<\delta_2$ \begin{align*}\frac{x^2+y^2}{\sin^2y +\ln\ (1+x^2)}  &=
\frac{x^2+y^2}{y^2 + O(y^3) + x^2 +O(x^3) } \\&<
\frac{x^2+y^2}{(1-\epsilon) x^2 + (1-\epsilon)y^2} \\&=
\frac{1}{1-\epsilon} \end{align*} for $0<\epsilon <1$, where $0<|x|,\ |y|<1\ \ast$
are small.
Similarly, we have $\frac{1}{1+\epsilon}<  \frac{x^2+y^2}{\sin^2y
+\ln\ (1+x^2)}$. In $\ast$, consider the case $x=0$ or $y=0$ so that
we have the limit $1$.
