Using L'Hopital Rule, evaluate $\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$

A single application of L'Hospital is sufficient:

$$\frac {1} {x^2}-\frac {\cot x} {x}=\frac{\sin x-x\cos x}{x^2\sin x}\xrightarrow{\text{L'Hospital}}\frac{x\sin x}{2x\sin x+x^2\cos x}=\frac{\dfrac{\sin x}x}{2\dfrac{\sin x}x+\cos x}\xrightarrow{\text{sinc}}\frac1{2+1}.$$

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.

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.