<s>[INST] 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!
 [/INST] What you have to use is the facts, for small $x,y$,
$$ \frac{|y|}{\sqrt2}\le|\sin y|\le |y|, \frac12{x^2}\le\ln (1+x^2)\le x^2$$
and
$$ \lim_{x\to0}\frac{x-\ln(1+x)}{x}=0,\frac{x-\sin x}{x}=0. $$
So
\begin{eqnarray*}
&&\bigg|\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}-1\bigg|\\
&=&\frac{x^2-\ln(1+x^2)+y^2-\sin^2y}{\sin^2y+\ln(1+x^2)}\\
&\le&2\frac{x^2-\ln(1+x^2)+y^2-\sin^2y}{x^2+y^2}\\
&=&2\frac{x^2-\ln(1+x^2)}{x^2+y^2}+2\frac{y^2-\sin^2y}{x^2+y^2}\\
&\le&2\frac{x^2-\ln(1+x^2)}{x^2}+2\frac{y^2-\sin^2y}{y^2}
&\to$0
\end{eqnarray*}
as $(x,y)\to(0,0)$.
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