## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{n\to\infty}\left(\cot\frac{1}{n}-n\right)=0.$$
To solve this limit we will introduce a substitution $t=\frac{1}{n}$. We see that when $n\to\infty$ then $t=\frac{1}{n}\to0^+$. LR will stand for Apply L'Hopital's rule: $$\lim_{n\to\infty}\left(\cot\frac{1}{n}-n\right)=\lim_{t\to0^+}\left(\cot t-\frac{1}{t}\right)=\lim_{t\to0^+}\left(\frac{\cos t}{\sin t}-\frac{1}{t}\right)=\lim_{t\to0^+}\frac{t\cos t-\sin t}{t\sin t}=\left[\frac{0\cdot\cos 0-\sin0}{0\cdot\sin0}\right]=\left[\frac{0}{0}\right][\text{LR}] = \lim_{t\to0^+}\frac{(t\cos t-\sin t)'}{(t\sin t)'}=\lim_{t\to0^+}\frac{(t)'\cos t+t(\cos t)'-(\sin t)'}{(t)'\sin t+t(\sin t)'}=\lim_{t\to0^+}\frac{\cos t-t\sin t-\cos t}{\sin t+t\cos t}=\lim_{t\to0^+}\frac{-t\sin t}{\sin t+t\cos t}=\left[\frac{-0^+\sin0^+}{\sin0^++0^+\cos0^+}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(-t\sin t)'}{(\sin t+t\cos t)'}=\lim_{t\to0^+}\frac{-(t)'\sin t-t(\sin t)'}{(\sin t)'+(t)'\cos t+t(\cos t)'}=\lim_{t\to0^+}\frac{-\sin t-t\cos t}{\cos t+\cos t-t\sin t}=\lim_{t\to0^+}\frac{-\sin t-t\cos t}{2\cos t-t\sin t}=\left[\frac{-\sin0^+-0^+\cos0^+}{2\cos0^+-0^+\sin0^+}\right]= \frac{-0-0\cdot1}{2\cdot 1-0\cdot0}=0.$$