Answer
$$\lim_{x\to\infty}x^{3/2}\sin(1/x)=\infty$$
Work Step by Step
$$A=\lim_{x\to\infty}x^{3/2}\sin(1/x)$$
$$A=\lim_{x\to\infty}\frac{x^{3/2}\sin(1/x)}{1}$$
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 $$A=\lim_{u\to0}\frac{\cos u}{3/2u^{1/2}}=\infty$$
