Does unbounded derivative imply No Lipschitz? We are taught that bounded derivative implies lipschitz. Also, at times, derivative may not exist, yet lipschitz could hold.

I wonder if derivative exists but is unbounded, can we directly say that it is not lipschitz?
Assume $f$ is Lipschitz and differentiable at $I$ and let
$a\in I$. 

$\exists K&gt;0: $

$\forall x\in I - \{a\}$

$|\frac{f(x)-f(a)}{x-a}| \leq K$

then by passage to the limit when $x\to a$, we get

$|f'(a)|\leq K$ and $f'$ is bounded at $ I$.
$$f(x) = x \sin\left(\frac1x\right) \text{ on } (0,1)$$ is a Lipschitz function with unbounded derivative