Check if function is Lipschitz continuous near 0

Question

I need to determine whether the function $f(x) = x^3\sin(1/x)$ is Lipschitz continuous near $x = 0$.

I observed that $|f'(x)| \rightarrow 0$ as $x \rightarrow 0$ and therefore $|f'(x)| < C$, where $C$ is some constant.

Now, using Lagrange formula $|f(x_1) - f(x_2)| \leq |f'(x)||x_1-x_2| < C|x_1-x_2|$, so $f(x)$ is Lipschitz continuous near $x = 0$.

Is my solution correct? If not, how should I solve it?

Answer 1

Yes, that looks correct.

(Note that "Lagrange formula" can mean a multitude of things, so it may be more accessible to say the Mean Value Theorem).