Limit $\frac{e^\frac{-x^2}{2}-\cos(x)}{x^3\sin(x)}$ as $x\to 0$ We have to find the limit:
$$\lim_{x\to 0}\dfrac{e^\frac{-x^2}{2}-\cos(x)}{x^3\sin(x)}$$
I was stuck, but was able to find the limit using series expansion as $\dfrac{1}{4}$.
How can we calculate the limit with standard limits like 
$$\lim_{x\to 0}\dfrac{e^x-1}{x}=1\\\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$$
etc.
Also I didn't try L'hospital as that would be too complicated.

We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e^{-x^{2}/2} - \cos x}{x^{3}\sin x}\notag\\
&= \lim_{x \to 0}\frac{e^{-x^{2}/2} - 1 + 1 - \cos x}{x^{4}}\cdot\frac{x}{\sin x}\notag\\
&= \lim_{x \to 0}\frac{e^{-x^{2}/2} - 1 + 2\sin^{2}(x/2)}{x^{4}}\notag\\
&= \lim_{t \to 0}\frac{e^{-2t^{2}} - 1 + 2\sin^{2}t}{16t^{4}}\text{ (putting }x = 2t)\notag\\
&= \frac{1}{16}\lim_{t \to 0}\frac{e^{-2t^{2}} - 1 + 2t^{2} + 2\sin^{2}t - 2t^{2}}{t^{4}}\notag\\
&= \frac{1}{16}\lim_{t \to 0}\frac{e^{-2t^{2}} - 1 + 2t^{2}}{t^{4}} + \frac{1}{8}\lim_{t \to 0}\frac{\sin^{2}t - t^{2}}{t^{4}}\notag\\
&= \frac{1}{4}\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}} + \frac{1}{8}\lim_{t \to 0}\frac{\sin t + t}{t}\cdot\frac{\sin t - t}{t^{3}}\text{ (putting }u = -2t^{2})\notag\\
&= \frac{1}{4}\cdot\frac{1}{2} - \frac{1}{8}\cdot 2\cdot\frac{1}{6}\notag\\
&= \frac{1}{12}\notag
\end{align}
The limits $$\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}} = \frac{1}{2},\lim_{t \to 0}\frac{\sin t - t}{t^{3}} = -\frac{1}{6}$$ are easily evaluated via Taylor's series or L'Hospital's Rule.
