if $f$ is differentiable at a point $x$, is $f$ also necessary lipshitz-continuous at $x$? if $f$ is differentiable at a point $x$, is $f$ also necessary Lipshitz at $x$?
Since $f$ is differentiable at $x$, $f$ is also continuous at $x$. Then we have the $\varepsilon$-$\delta$ definition of $f$ continuous at $x$, and also we have $f'(x)$ exists. But i have no idea how to connect them together. Since $f$ is only differentiable at a point, i don't think mean value theorem is gonna work here either.
 A: If $f$ is differentiable at $x_0$, then for all $\epsilon>0$, there exists a $\delta>0$ such that if $\|x-x_0\|< \delta$, then $\|f(x)-f(x_0)-DF(x_0)(x-x_0)\| \leq \epsilon \|x-x_0\|$. S, choose $\epsilon=1$, then the estimate gives $\|f(x)-f(x_0)\|-\|DF(x_0)(x-x_0)\| \leq \|x-x_0\|$, which can be written as $\|f(x)-f(x_0)\| \leq (1+\|Df(x_0)\|) \|x-x_0\|$. Setting $L = 1+\|Df(x_0)\|$ shows that $\|f(x)-f(x_0)\| \leq L \|x-x_0\|$ locally.
However, being differentiable at a point does not mean that $f$ is locally Lipschitz. For example, let $f(x) = x^2 \sin \frac{1}{x^2}$. $f$ is differentiable at $x=0$, but is not locally Lipschitz around $x=0$ ($f'(x)$ is unbounded near $x=0$). 
If the function is differentiable, then it is locally Lipschitz in a neighborhood iff the derivative is bounded on the neighborhood .
A: Lipschitz continuity is stronger than mere continuity. If the function $f$ is continuously differentiable at $x_0$, you can find a Lipschitz constant using a Taylor series or a mean value theorem for the derivative which will assume its finite maximum on a small compact interval around $x_0$ (which will be the local Lipschitz constant). The same holds for any function for which the derivative is locally bounded near $x_0$. But there are functions which are differentiable at a point but not Lipshitz continuous. I blanked on an example, and stole this from Wikipedia - a differentiable function on the compact set $[0, 1]$ that is not Lipshitz: 
$$f(x) = x^{\frac{3}{2}} \sin(\frac{1}{x}), \, x ≠ 0; \quad  f(0) = 0$$
