## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{n\to\infty}\left(n\cot\frac{1}{n}-n^2\right)=-\frac{1}{3}.$$
To solve this limit we will introduce the substitution $t=\frac{1}{n}$. In this case when $n\to\infty$ then $t=\frac{1}{n}\to0^+$. LR will stand for Apply L'Hopital's rule. $$\lim_{n\to\infty}\left(n\cot\frac{1}{n}-n^2\right)=\lim_{t\to0^+}\left(\frac{1}{t}\cot t-\frac{1}{t^2}\right)=\lim_{t\to0^+}\frac{t\cot t-1}{t^2}=\lim_{t\to0^+}\frac{t\cos t-\sin t}{t^2\sin t}=\left[\frac{0^+\cos0^+-\sin0^+}{(0^+)^2\sin0^+}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(t\cos t-\sin t)'}{(t^2\sin t)'}=\lim_{t\to0^+}\frac{(t)'\cos t+t(\cos t)'-(\sin t)'}{(t^2)'\sin t+t^2(\sin t)'}=\lim_{t\to0^+}\frac{\cos t-t\sin t-\cos t}{2t\sin t+t^2\cos t}=\lim_{t\to0^+}\frac{-t\sin t}{t(2\sin t+t\cos t)}=\lim_{t\to0^+}\frac{-\sin t}{2\sin t+t\cos t}=\left[\frac{-\sin0^+}{2\sin0^++0^+\cos 0^+}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(-\sin t)'}{(2\sin t+t\cos t)'}=\lim_{t\to0^+}\frac{-\cos t}{2\cos t+(t)'\cos t+t(\cos t)'}=\lim_{t\to0^+}\frac{-\cos t}{2\cos t+\cos t-t\sin t}=\lim_{t\to0^+}\frac{-\cos t}{3\cos t-t\sin t}=\frac{-\cos 0^+}{3\cos0^++0^+\sin0^+}=-\frac{1}{3}.$$