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.

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.
