How to find a function that is locally Lipschitz, but not Lipschitz overall?

Question

How to find a function that is locally Lipschitz, but not Lipschitz overall?

This is question for my homework, but I don't understand how that function can exist.

Answer 1

You can just take $x^2$. It is Lipschitz on every closed compact set but, as you allow the closed compact sets to get larger and larger, the Lipschitz constant grows to infinity, such that the function is not Lipschitz 'overall'.

Answer 2

Please take a look at the comments. As these two hints are sufficient for the answer, I'll combine them here for you.

Hint 1 from uniquesolution: Every continuously differentiable function is locally Lipschitz

Hint 2 from Sangchul Lee: Every continuously-differentiable and Lipschitz function on an interval has bounded derivative.

Now, consider the following function \begin{equation} f(x)=\sin{(x^2)}. \end{equation} Does this function satisfy the first hint? Can you compute the derivative to see that the second hint 'fails'? If this does not clarify the question, let me know in the comments and I will elaborate more.