Calculate the limit of the function Please help me to calculate the limit of the function. I do not know where to start.

Thank you.
 A: Let 
$$
f(x)= \left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]
$$
and $$g(x)=-\frac{1}{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}.$$
We have to find the limit for $x\to0$ of the function $\varphi(x)=[f(x)]^{-\frac{1}{g(x)}}$
$$
\lim_{x\to 0}\varphi(x)=\lim_{x\to 0}[f(x)]^{g(x)}
$$
For $x\to 0$
for the first term in the bracket we have
$$
\begin{align}
\sin(x+x^2)&\sim x+x^2-\frac{(x+x^2)^3}{6}\sim x+x^2-\frac{x^3}{6}-\frac{x^4}{2}+o(x^4)\\
%
\frac{1}{1-x^2}&\sim 1+x^2+x^4+o(x^4)\\
%
\frac{\sin(x+x^2)}{1-x^2}&\sim \left(x+x^2-\frac{x^3}{6}-\frac{x^4}{2}\right)\left( 1+x^2+x^4\right)\sim x+x^2+\frac{5}{6}x^3+\frac{x^4}{2}+o(x^4)\\
\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2&\sim \left(x+x^2+\frac{5}{6}x^3+\frac{x^4}{2}\right)\sim x+2x^2+\frac{8}{3}x^4+o(x^4)
\end{align}$$
For the second term
$$
\begin{align}
\arctan(x^2)&\sim x^2+x^4+o(x^4)\\
\frac{\arctan(x^2)}{1-x^2}&\sim \left(x^2+x^4\right)\left( 1+x^2+x^4\right)\sim x^2+2x^4+o(x^4)\\
\end{align}
$$
So the Taylor expansion for $f(x)$ at the order 4 is
$$
f(x)\sim x+2x^2+\frac{8}{3}x^4-(x^2+2x^4)-2x^3+1\sim 1+\frac{2}{3}x^4+o(x^4)
$$
Now let find the Taylor expansion for $g(x)$. Observing that
$$
\begin{align}
1-\operatorname{e}^{2x^2}&\sim 1-\left(1-2x^2-\frac{(2x)^2}{2}\right)=-2x^2+2x^4+o(x^4)\\
\sinh(3x^2)&\sim 3x^2+\frac{(3x^2)^3}{6}\sim 3x^2+o(x^4)
\end{align}
$$
we find
$$
g(x)\sim-\frac{1}{6x^4+o(x^4)}
$$
Putting all together we have
$$
\varphi(x)
\sim\left[1+\frac{2}{3}x^4\right]^{-1/(6x^4)}=\exp\left({-\frac{1}{6x^4}}\log\left(1+\frac{2}{3}x^4\right)\right)
$$
and observing that
$$
\log\left(1+\frac{2}{3}x^4\right)\sim \frac{2}{3}x^4+o(x^4)
$$
we finally find for $x\to 0$
$$
\varphi(x)\sim\exp\left(-\frac{1}{6x^4}\frac{2}{3}x^4\right)\to \operatorname{e}^{-1/9}.
$$
A: To get started, Assuming limit L exists then 
$$
 \begin{align} 
\\ \lim_{x\to0} f(x)^{g(x)} &={\rm L}
\\ \lim_{x\to0}  {g(x)} \ln f(x)  &=\ln {\rm L}
\end{align}
$$
Where 
$$
 \begin{align} 
f(x)&= \left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]
 \\ g(x) &=-\frac{1}{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}.
\end{align}
$$
so $$
 \begin{align} 
\\ \lim_{x\to0}  {g(x)} \ln f(x)  &=\ln {\rm L}
\\ \lim_{x\to0} -\frac{\ln\left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]}{{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}}&= \ln {\rm L}
\\ \lim_{x\to0} \frac{\ln\left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]}{{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}}&= \ln {\rm \frac{1}{L}}
\end{align}
$$
The last limit is of the form $\frac{0}{0}$, so appeal to sprite of L'Hopital
$$\begin{align}\\ \lim_{x\to0} \frac{\frac{\rm d}{{\rm d}x}\left(\ln\left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]\right)}{{\frac{\rm d}{{\rm d}x}\left(\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)\right)}}&= \ln {\rm \frac{1}{L}}
\end{align}
$$
Now either the above can be evaluated, if not repeat L'Hopital for desired result.
