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!
 A: 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 $(x,y)\to(0,0)$.
A: 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 $1$.
A: You can always pass to polar coordinates to compute 2d limits:
Set $x=r \cos\theta , y=r\sin \theta$, and now we need to compute
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)}$$ 
Since this is a limit for the variable $r$, you can use L'Hôpital taking derivatives w.r.t. $r$. Note that if we end with something depending on $\theta$, the limit does not exist, becasuse the limit (if exists) is unique.
After L'Hôpital and some basic manipulation, we end with
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)} = \lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{\sin (2r\sin\theta) \sin \theta(1+r^2\cos^2\theta) + 2r\cos^2\theta}$$
You use the always useful $\sin (ar) \approx ar$ when $r \approx 0$ to rewrite the denominator of the RHS:
$$\lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{(2r\sin\theta)\sin\theta(1+r^2\cos^2\theta) + 2r\cos^2\theta} = \lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{2r\sin^2 \theta(1+r^2\cos^2\theta) + 2r\cos^2\theta}= \dfrac{1}{\sin^2 \theta + \cos^2\theta} = 1$$
A: Swapping numerator and denominator,
$$
\frac{\sin^2 y+\ln(1+x^2)}{x^2+y^2}=\frac{y^2}{x^2+y^2}\frac{\sin^2 y}{y^2}+\frac{x^2}{x^2+y^2}\frac{\ln(1+x^2)}{x^2}.
$$ It is equal to
$$
\frac{y^2}{x^2+y^2}\left(\frac{\sin^2 y}{y^2}-1\right)+\frac{x^2}{x^2+y^2}\left(\frac{\ln(1+x^2)}{x^2}-1\right)+1
$$ The first term tends to $0$ as
$$
\left|\frac{y^2}{x^2+y^2}\left(\frac{\sin^2 y}{y^2}-1\right)\right|\le \left|\frac{\sin^2 y}{y^2}-1\right|\xrightarrow{x,y\to 0} 0.
$$ Similarly for the second term
$$\left|\frac{x^2}{x^2+y^2}\left(\frac{\ln(1+x^2)}{x^2}-1\right)\right|\le \left|\frac{\ln(1+x^2)}{x^2}-1\right|\xrightarrow{x,y\to 0} 0.
$$ So, it follows
$$
\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sin^2 y+\ln(1+x^2)}=\lim_{(x,y)\to (0,0)}\frac1{\frac{\sin^2 y+\ln(1+x^2)}{x^2+y^2}}=1.
$$
A: We have $\sin^2 y=y^2(1+f(y))$ and $\ln (1+x^2)=x^2(1+g(x))$ where $\lim_{y\to 0}f(y)=0=\lim_{x\to 0}g(x).$
So $\lim_{x^2+y^2\to 0}|f(y)|+|g(x)|=0.$
When $x^2+y^2\ne 0,$ the reciprocal of the expression in the Q is $1+\frac {y^2f(y)+x^2g(x)}{x^2+y^2}$ which cannot differ from $1$ by more than $\frac {(y^2+x^2)(|f(y)|+|g(x)|)}{x^2+y^2}=|f(y)|+|g(x)|.$
