What is wrong with the calculation $ \lim\limits_{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $? We are given
$$ \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $$
We can write
$$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} = \lim _{x \to 0} \frac {x (cos x - \sin x/x)} {x^2 \sin x} $$
It can be written as
$$ L = \lim _{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} $$
We know
$$ \lim _{x \to 0} \frac {\sin x} {x} = 1 $$
The original limit can be written as
$$ L = \lim _{x \to 0} \frac { \cos x - 1} {x^2} $$
Now using $ ( \cos x ) $ expansion formula, the answer will be $ L = -\frac12$ , but the answer is $ L = -\frac13$ , which is not matching.
I want to know what I did wrong in this method. I know the answer using other methods, but I want to know what went wrong with this one.

The thing is you can take a partial limit when it only involves products (or quotients). Here there is an additive $\cos(x)$ term so you cannot do that.
Remember that $\cos(x) = 1 - \frac{x^2}{2} + o(x^2)$ and $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ therefore $\cos x - \frac{\sin x}{x} = -\frac{x^2}{3} + o(x^2)$ which means that $ \frac { \cos x - \sin x/x} {x^2 }\to -\frac{1}{3}. $
Therefore $$\lim_{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} = \lim_{x \to 0} \frac { -\frac{1}{3} } {\sin x/x} = -\frac{1}{3}  $$
