## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to\infty}x^3\left(\frac{1}{x}-\sin \frac{1}{x}\right)=\frac{1}{6}.$$
To solve this limit we will first introduce a substitution $t=\frac{1}{x}.$ In that case also $x=\frac{1}{t}$ and when $x\to\infty$ then $t\to0^+$. Also, we will use L'Hopital's rule. LR will stand for Apply L'Hopital's rule $$\lim_{x\to\infty} x^3\left(\frac{1}{x}-\sin\frac{1}{x}\right)=\lim_{t\to0^+}\left(\frac{1}{t}\right)^3(t-\sin t)=\lim_{t\to0^+}\frac{t-\sin t}{t^3}=\left[\frac{0^+-\sin0^+}{(0^+)^3}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(t-\sin t)'}{(t^3)'}=\lim_{t\to0^+}\frac{1-\cos t}{3t^2}=\left[\frac{1-\cos 0^+}{3(0^+)^3}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(1-\cos t)'}{(3t^2)'}=\lim_{t\to0^+}\frac{\sin t}{6t}=\frac{1}{6}\lim_{x\to0^+}\frac{\sin t}{t}=\frac{1}{6}\cdot 1=\frac{1}{6},$$ where in the last step we used the fact that $\lim_{t\to0}\frac{\sin t}{t}=1$ which is a known ready-to-use result.