How to calculate $\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$? I need some help with computing this limit:
$$\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$$
I'm guessing that I should do a Taylor expansion of $\cos(\sin x)$ in base, but what should I do with the stuff in the exponent?

This answer is a codicil to that of Paramanand Singh. I give here a different calculation of $$\lim\limits_{x\to 0} \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}$$ 
Frequent use of the replacement of $\sin y$ by $\frac{\sin y}{y}y$ is made.
We have 
$$ \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}= \frac{\cos(\sin x)-\cos x}{x^4}
+ \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}.$$ And we calculate each separately.
For the first we use the difference of cosines formula $\cos a-\cos b=2\sin \frac{a+b}{2}\sin \frac{b-a}{2}$
So we have $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin \frac{\sin x+x}{2}}{x}\frac{\sin \frac{x-\sin x}{2}}{x^3}$$
the standard replacement above reduces to the limit of 
$$\frac{1}{2}\frac{\sin x+x}{x}\frac{x-\sin x}{x^3}\rightarrow \frac{1}{6}.$$
As for the limit $\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$, many people would be willing to take this as granted but I give here a nice elementary derivation.
First $$\left(\frac{1-\cos x}{x^2}\right)^2=\frac{2-2\cos x-\sin^2 x}{x^4}\rightarrow \frac{1}{4}$$
Second $$\frac{x-\sin x}{x^3}\frac{\sin x}{x}=\frac{x\sin x-\sin^2 x}{x^4}\rightarrow \frac{1}{6}$$
Subtracting the first from the second gives
$$\frac{2\cos x-2 +x\sin x }{x^4}\rightarrow -\frac{1}{12} $$
we can rewrite this as
$$2\frac{\cos x-1 +\frac{x^2}{2} }{x^4}+\frac{\sin x -x}{x^3}\rightarrow -\frac{1}{12} $$ and this gives 
$$\frac{\cos x-1 +\frac{x^2}{2} }{x^4}\rightarrow \frac{1}{24} $$
Finally adding them together we get 
$$\frac{1}{6}+\frac{1}{24}=\frac{5}{24}.$$
