## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}x^{3/2}\sin(1/x)=\infty$$
$$A=\lim_{x\to\infty}x^{3/2}\sin(1/x)$$ $$A=\lim_{x\to\infty}\frac{x^{3/2}\sin(1/x)}{1}$$ Divide both numerator and denominatory by $x^{3/2}$, and we have $$A=\lim_{x\to\infty}\frac{\sin(1/x)}{\frac{1}{x^{3/2}}}$$ $$A=\lim_{x\to\infty}\frac{\sin(1/x)}{(\frac{1}{x})^{3/2}}$$ Let's take $u=\frac{1}{x}$, which means $u$ approaches $0$ when $x$ approaches $\infty$. $$A=\lim_{u\to0}\frac{\sin u}{u^{3/2}}$$ Since $\lim_{u\to0}(\sin u)=\sin0=0$ and $\lim_{u\to0}(u^{3/2})=0^{3/2}=0$, this is an indeterminate form of $\frac{0}{0}$. Using L'Hospital's Rule: $$A=\lim_{u\to0}\frac{\cos u}{(3/2)u^{1/2}}$$ We have $\lim_{u\to0}(\cos u)=\cos0=1$ $\lim_{u\to0}(\frac{3}{2}u^{1/2})=\frac{3}{2}\times0^{1/2}=0$ That means $$A=\lim_{u\to0}\frac{\cos u}{3/2u^{1/2}}=\infty$$