# Locally Lipschitz

1. Feb 20, 2008

### gaganaut

Would a trig function like $$tan \left(x\right)$$ be locally Lipschitz?

How do we know that, if we know that $$tan \left(x\right)$$ is not continuously differentiable?

Last edited: Feb 20, 2008
2. Feb 20, 2008

### quasar987

tan(x) is continuously differentiable everywhere where it is defined.

And following my geometrical intuition, I would say that it is locally lip****z, and that you would have to try hard to find a function that is continuous but not locally lipschitz!

3. Feb 1, 2011

### Landau

Not that hard though, e.g.
$$[-1,1]\to\mathbb{R}$$
$$x\mapsto x^{1/3}$$
is not Lipschitz on any nhbd of zero.