Answer
$$0$$
Work Step by Step
We have
$$
\lim _{x \rightarrow 0} \left(\cot x -\frac{1}{x}\right) =\infty-\infty
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0} \frac{x\cot x -1}{x}=\lim _{x \rightarrow 0} \frac{\cot x-x\csc^2 x }{1}=-0\times\infty+\infty
.$$
Again we can apply L’Hôpital’s Rule
$$
\lim _{x \rightarrow 0} ( \cot x-x\csc^2 x) = \lim _{x \rightarrow 0} \frac{\cos x \sin x-x}{\sin^2 x} = \lim _{x \rightarrow 0} \frac{- \sin^2 x+\cos^2 x-1}{2\sin x\cos x} =0
.$$
