The function $sin(x^2)$ is not Lipschitz continuous on $\mathbb{R}?$
My steps are like this:
Suppose assume for some $C\geq 0$ $$ \bigg|\frac{sin(x^2)-sin(y^2)}{y-x}\bigg|\leq C$$ is true, for all $y\neq x,$ where $x, y\in \mathbb{R}.$
Now how we should proceed further?
It is even not uniformly continuous on the real line: Take $x_{n}=\sqrt{2n\pi+\pi/2}$ and $y_{n}=\sqrt{2n\pi}$, then $x_{n}-y_{n}\rightarrow 0$ but $\sin(x_{n}^{2})-\sin(y_{n}^{2})=1$ does not converge to $0$ as $n\rightarrow\infty$.
If it were Lipschitz continuous, then it is uniformly continuous.