calculus - Does the limit $limlimits_{x toinfty}x^2 sinfrac{1}{x}$ 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?