<s>[INST] 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.
 [/INST] Note that
$$
\lim_{x\to0}x\cot x=\lim_{x\to0}\frac{x}{\sin x}\cos x=1
$$
so $\dfrac{1-x\cot x}{x^2}$ is an indeterminate form $0/0$ at $0$.
You can certainly use l’Hôpital:
$$
\lim_{x\to0}\dfrac{1-x\cot x}{x^2}
=
\lim_{x\to0}\frac{-\cot x+\frac{x}{\sin^2x}}{2x}=
\lim_{x\to0}\frac{x-\sin x\cos x}{2x\sin^2x}
$$
However, this doesn't seem very inviting, but not hard at all. Observe that you can compute instead
$$
\lim_{x\to0}\frac{x-\sin x\cos x}{2x^3}=
\lim_{x\to0}\frac{1-\cos^2x+\sin^2x}{6x^2}=
\lim_{x\to0}\frac{2\sin^2x}{6x^2}
$$
Alternatively you can do
$$
\frac{1}{x^2}-\frac{\cot x}{x}=
\frac{1}{x^2}-\frac{\cos x}{x\sin x}=
\frac{\sin x-x\cos x}{x^2\sin x}
$$
which is much nicer:
$$
\lim_{x\to0}\frac{\sin x-x\cos x}{x^2\sin x}=
\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}\frac{x}{\sin x}
$$
Since the limit of the second fraction is $1$, we can just compute
$$
\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=
\lim_{x\to0}\frac{x\sin x}{3x^2}
$$
which is fairly easy.
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