Using L'Hopital Rule, evaluate $\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$ Using L'Hopital Rule, evaluate $$ \lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$$ 
I find this question weired.
If we just combine the two terms into one single fraction, we get$$\lim_{x \to 0} {\frac {1-x\cot x} {x^2}}=\frac10=\infty$$ 
If we follow L'Hopital Rule, this is $\infty-\infty$ form. We follow the following process to convert it into $\frac00$form.$$\infty_1 -\infty_2=\frac 1{\frac 1{\infty_1}}-\frac 1{\frac 1{\infty_2}}=\frac {\frac 1{\infty_2}-\frac 1{\infty_1}}{{\frac 1{\infty_1}}{\frac 1{\infty_2}}}$$
So we will get $$\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}=\lim_{x \to 0} {\frac {x\tan x-x^2}{x^3\tan x}}$$
If you keep differentiating using the rule you will get rid of the form of $\frac00$ in the third step of differentiation, which give you the answer $1 \over 3$. This method is very tedious. Trust me, you don't want to try.
I am wondering is there a smarter way of solving this question?
Thanks.

first : 
$$\lim_{x\to 0} \frac{\sin x-x}{x^3}=\frac{-1}{6} \tag{1}$$
$$\lim_{x\to 0} \frac{1-\cos x}{x^2}=\frac{3}{6} \tag{2}$$
$$\lim_{x\to 0} \frac{x}{\sin x}=1 \tag{3} $$
$$\frac{1}{x^2}-\frac{\cot x}{x}=\frac{\sin x-x}{x^3}\times \frac{x}{\sin x}+\frac{1-\cos x}{x^2}\times \frac{x}{\sin x}$$
$$L=\lim_{x\to 0} \frac{1}{x^2}-\frac{\cot x}{x}=\lim_{x\to 0}(\frac{\sin x-x}{x^3}\times \frac{x}{\sin x}+\frac{1-\cos x}{x^2}\times \frac{x}{\sin x})$$
Now using $(1) ,(2),(3)$:
$$L=(\frac{-1}{6}\times 1+\frac{3}{6}\times 1)=\frac{2}{6}$$
