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.
 A: If you do not want to use L'Hospital, I suppose that Taylor expansions are the good way to go.
Considering 
$$y=\frac {x\tan( x)-x^2}{x^3\tan (x)}$$ and $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ we then have $$y=\frac{\frac{x^4}{3}+\frac{2 x^6}{15}+O\left(x^7\right)}{x^4+\frac{x^6}{3}+\frac{2 x^8}{15}+O\left(x^9\right)}=\frac{1}{3}+\frac{x^2}{45}+O\left(x^3\right)$$ which shows the limit and also how it is approached.
A: 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.
A: A single application of L'Hospital is sufficient:
$$\frac {1} {x^2}-\frac {\cot x} {x}=\frac{\sin x-x\cos x}{x^2\sin x}\xrightarrow{\text{L'Hospital}}\frac{x\sin x}{2x\sin x+x^2\cos x}=\frac{\dfrac{\sin x}x}{2\dfrac{\sin x}x+\cos x}\xrightarrow{\text{sinc}}\frac1{2+1}.$$
A: I would keep the fraction as $$\frac{1-x\cot x}{x^2} $$
and use the fact that $(\cot)'(x) = -1-\cot^2x$. In this way the second derivative of $1-x \cot x$ is not too bad to calculate at all.
A: A Laurent series is a Taylor expansion with negative exponents. The Laurent expansion for $\cot(x)$ is
$$\cot(x) \approx \frac{1}{x} - \frac{x}{3} - O(x^3) \\ 
\lim_{x \to 0} \left(\frac{1}{x^2} - \frac{\cot(x)}{x}\right) \rightarrow \frac{1}{x^2} - \frac{1}{x^2} + \frac{1}{3} + O(x^2) = \frac{1}{3}$$
A: 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 $(1) ,(2),(3)$:
$$L=(\frac{-1}{6}\times 1+\frac{3}{6}\times 1)=\frac{2}{6}$$
