Compute $\lim_{x\rightarrow0}\frac{e^{x^2} - \cos x}{\sin^2 x}$ Find the limit as $x$ approaches $0$ of 
$\dfrac{e^{x^2} - \cos x}{\sin^2 x}$
What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\frac 12$.
Is this right? Any help would be appreciated.

Use Taylor expansion. As $x\rightarrow0$, you have
$$e^{x^2}=1+x^2+O(x^4)$$
$$\cos x=1-\frac12x^2+O(x^4)$$
$$\sin^2x=x^2+O(x^4)$$
Thus
$$\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{1+x^2-1+\frac12x^2+O(x^4)}{x^2+O(x^4)}=\frac{\frac32+O(x^2)}{1+O(x^2)}\underset{x\rightarrow 0}{\longrightarrow} \frac32$$
