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.
 A: Using Taylor polynomials, you have 
$$
\frac {1-\frac {x^2}2+\frac {x^4}{8}+O (x^6)-(1-\frac {x^2}2+\frac {x^4}{24}+O (x^6))}{x^3\sin x}
= \frac {\frac {x^4}{12}+O (x^6)}{x^3\sin x}\to\frac1 {12}.
$$
You cannot expect to use limits as simple as those in your question, because this limit depends on the terms of degree two in the expansion, while the two limits you quote depend on the terms of  degree one.
A: 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.
A: Try this
add 1 and minus 1 from the numerator.
take the exponent along with -1 and the cos function along with +1
now apply basic limit (like the property u have stated above for exponents) and expand $$ 1-cos(x)$$ to get $$2sin^2(\frac{x}{2})$$ 
U will reach the answer. (Apologies. Newbie at LaTeX)
