## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{n\to\infty}n^2\ln\left(n\sin\frac{1}{n}\right)=-\frac{1}{6}.$$
To solve this limit we will introduce a substitution $t=\frac{1}{n}$. In that case when $n\to\infty$ then $t=\frac{1}{n}\to0^+$. LR will stand for Apply L'Hopital's rule. $$\lim_{n\to\infty}n^2\ln\left(n\sin\frac{1}{n}\right)=\lim_{t\to0^+}\frac{\ln\left(\frac{\sin t}{t}\right)}{t^2}=\left[\frac{\ln\lim_{t\to0^+}\frac{\sin t}{t}}{(0^+)^2}\right]=\left[\frac{\ln 1}{0^+}\right]=\left[\frac{0}{0}\right][\text{LR}]= \lim_{t\to0^+}\frac{\left(\ln\left(\frac{\sin t}{t}\right)\right)'}{(t^2)'}=\lim_{t\to0^+}\frac{\frac{1}{\frac{\sin t}{t}}\left(\frac{\sin t}{t}\right)'}{2t}=\lim_{t\to0^+}\frac{\frac{t}{\sin t}\frac{(\sin t)'t-\sin t(t)'}{t^2}}{2t}=\lim_{t\to0^+}\frac{t\cos t-\sin t}{2t^2\sin t}=\left[\frac{0^+\cos0^+-\sin0^+}{2(0^+)^2\sin0^+}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(t\cos t-\sin t)'}{(2t^2\sin t)'}=\lim_{t\to0^+}\frac{(t)'\cos t+t(\cos t)'-\cos t}{2(t^2)'\sin t+2t^2(\sin t)'}=\lim_{t\to0^+}\frac{\cos t-t\sin t-\cos t}{4t\sin t+2t^2\cos t}=\lim_{t\to0^+}\frac{-t\sin t}{2t(2\sin t+t\cos t)}=\lim_{t\to0^+}\frac{-\sin t}{2(2\sin t+t\cos t)}=\left[\frac{-\sin0^+}{2(\sin0^++0^+\cos0^+)}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(-\sin t)'}{(2(2\sin t + t\cos t))'}=\lim_{t\to0^+}\frac{-\cos t}{2(2\cos t+(t)'\cos t+t(\cos t)')}=\lim_{t\to0^+}\frac{-\cos t}{2(2\cos t+\cos t-t\sin t)}=\frac{-\cos 0}{2(2\cos0+\cos0-0\cdot\sin0)}=\frac{-1}{2(2+1-0)}=-\frac{1}{6}.$$