Answer
The solution is
$$\lim_{x\to0^+}\left(\cot x-\frac{1}{x}\right)=0.$$
Work Step by Step
To calculate this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to0^+}\left(\cot x-\frac{1}{x}\right)=\lim_{x\to0^+}\left(\frac{\cos x}{\sin x}-\frac{1}{x}\right)=\lim_{x\to0^+}\frac{x\cos x-\sin x}{x\sin x}=\left[\frac{0^+\cos0^+-\sin0^+}{0^+\sin0^+}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0^+}\frac{(x\cos x-\sin x)'}{(x\sin x)'}=\lim_{x\to0^+}\frac{(x)'\cos x+x(\cos x)'-\cos x}{(x)'\sin x+x(\sin x)'}=\lim_{x\to0^+}\frac{\cos x-x\sin x-\cos x}{\sin x+x\cos x}=\lim_{x\to0^+}\frac{-x\sin x}{\sin x+x\cos x}=\left[\frac{-0^+\cdot\sin0^+}{\sin0^+ + 0^+\cdot\cos0^+}\right]=\frac{0}{1}=0.$$
