Finding $\lim\limits_{x\to\infty}x^{3/2}\cdot\sin\frac1x$ How would I go about solving this problem? Not really sure where to start.
Put $y = \dfrac{1}{x} \implies x = \dfrac{1}{y}\implies L = \displaystyle \lim_{y \to 0} \dfrac{\sin y}{y\sqrt{y}}= \infty$
Let $h=\frac{1}{x}$

As $x\to \infty ,h\to 0$.

So $x^{\frac{3}{2}}{\sin{\frac{1}{x}}}=\dfrac{\sin h}{h^{\frac{3}{2}}}$

Put $h^{\frac{3}{2}}=t\implies h=t^{\frac{2}{3}}$

So $\dfrac{\sin h}{h^{\frac{3}{2}}}=\dfrac{\sin t^{\frac{2}{3}}}{t}=(\dfrac{0}{0})\text{as} t\to 0$

Applying L'Hospital's Rule ,we get $\dfrac{\cos t^{\frac{2}{3}}\times\frac{2}{3}}{t^{\frac{1}{3}}}$

which tends to $-\infty $ as $t\to 0^{-}$ and tends to $\infty $ as $t\to 0^{+}$

So the limit does not exist.