Question on Lipschitz continuity

Question

How to check whether the function $$f(x)=x\sin\bigg(\frac{1}{1+x^2}\bigg)$$ is Lipschitz or not on $\mathbb{R}?$ I tried by considering $$\bigg|\frac{x \sin\Big(\frac{1}{1+x^2}\Big)-y \sin\Big(\frac{1}{1+y^2}\Big)}{x-y} \bigg|\leq C.$$ Then how to proceed further?

Answer 1

using the mean value Theorem we get $$|f(x)-f(y)|=\left|\frac{-2\xi^2}{(1+\xi^2)^2}\cos\left(\frac{1}{1+\xi^2}\right)+\sin\left(\frac{1}{1+\xi^2}\right)\right||x-y|$$ and $$f'(\xi)$$ is bounded

Answer 2

Given that $\frac{2x^2}{(1+x^2)^2}=\frac{x^2}{(1+x^2)}\frac{1}{(1+x^2)}\le 2$ we have, $$\left|f'(x)\right|= \left|\sin\left(\frac{1}{1+x^2}\right)-\frac{2x^2}{(1+x^2)^2}\cos\left(\frac{1}{1+x^2}\right)\right|\\\le\left|\sin\left(\frac{1}{1+x^2}\right)\right|+\left|\frac{2x^2}{(1+x^2)^2}\cos\left(\frac{1}{1+x^2}\right)\right|\le 3$$

is bounded then it is Lipschitz with constant $3$.