indeterminate forms - Limit $frac{e^frac{-x^2}{2}-cos(x)}{x^3sin(x)}$ as $xto 0$

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.

Answer

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.