Does the limit $\lim\limits_{x \to\infty}x^2 \sin\frac{1}{x}$ not exist? I used $u$ substitution for the limit $$\lim_{x \to\infty}x^2 \sin\frac{1}{x}$$ and got the limit does not exist by saying $u=\frac{1}{x}$. Is this correct and if so would that mean $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x^2}$ and $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x}$ also don't exist?

Since $\sin\frac{1}{x}\to0$ and $x^2\to\infty$, we use l'Hôpital's theorem:
$$\lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x^2}}=\lim_{x\to\infty}\frac{-\frac{1}{x^2}\cos\frac{1}{x}}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{x}{2}\cos\frac{1}{x}=\infty$$
since $\cos\frac{1}{x}\to1$ and $\frac{x}{2}\to\infty$, as $x \rightarrow \infty.$
