## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to\infty}x^2\ln\left(\cos\frac{1}{x}\right)=-\frac{1}{2}.$$
To solve this limit we will use L'Hopital's rule. LR will stand for Apply L'Hopital's rule: $$\lim_{x\to\infty}x^2\ln\left(\cos\frac{1}{x}\right)=\left[\infty^2\ln\cos\frac{1}{\infty}\right]=\left[\infty\ln\cos 0\right]=[\infty\cdot0]=\lim_{x\to\infty}\frac{\ln\left(\cos\frac{1}{x}\right)}{\frac{1}{x^2}}=\left[\frac{\ln\cos1/\infty}{1/\infty^2}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{\left(\ln\left(\cos\frac{1}{x}\right)\right)'}{\left(\frac{1}{x^2}\right)'}=\lim_{x\to\infty}\frac{\frac{1}{\cos\frac{1}{x}}\left(\cos\frac{1}{x}\right)'}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{\frac{1}{\cos\frac{1}{x}}(-\sin\frac{1}{x})\left(\frac{1}{x}\right)'}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{\frac{1}{x^2}\tan\frac{1}{x}}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{-\tan\frac{1}{x}}{\frac{2}{x}}=\left[\frac{-\tan1/\infty}{2/\infty}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{\left(-\tan\frac{1}{x}\right)'}{\left(\frac{2}{x}\right)'}=\lim_{x\to\infty}\frac{-\frac{1}{\cos^2\frac{1}{x}}\left(\frac{1}{x}\right)'}{2\left(\frac{1}{x}\right)'}=\lim_{x\to\infty}\frac{-1}{2\cos^2 \frac{1}{x}}=\left[\frac{-1}{2\cos1/\infty}\right]=\frac{-1}{2\cos 0}=-\frac{1}{2}.$$