Lipschitz continuous and differentiability Hey i'm trying to solve this question:
Let U be an open interval in R and a ∈ U. 
Prove that if f : U → R is Lipschitz
continuous, then $g(x) = (f(x) − f(a))^2$
is differentiable at a.
so far I have the following, but im not sure how to piece it all together
$|f(x) - f(y)| \le k|x-y|$
=> $ (f(x) − f(a))^2 \le (k|x-a|)^2$
if $g(x)$ is differentiable at a then $\lim_{x\to a} \frac{g(x) - g(a)}{x - a }$ must exist
$\lim_{x\to a} \frac{g(x) - g(a)}{x - a }  = \lim_{x\to a} \frac{(f(x) − f(a))^2 - (f(a) − f(a))^2}{x - a } = \lim_{x\to a} \frac{(f(x) − f(a))^2}{x - a } \le \frac{(k|x-a|)^2}{x-a} = k^2(x-a)$
and from here im not sure where to go to show the limit exists
 A: hint
For $x\ne a$,
$$|\frac{g(x)-g(a)}{x-a}|=$$
$$|\frac{f(x)-f(a)}{x-a}||f(x)-f(a)|$$
$$\le K|f(x)-f(a)|$$$f$ is continuous at $a$ then
$$\lim_{x\to a}(f(x)-f(a))=0$$
then
$$g'(a)=0.$$
A: To prove that, we find the left and write limits.$$\lim_{x\to a^+} \frac{g(x) - g(a)}{x - a }=\lim_{x\to a^+} \frac{g(x)}{|x - a|}$$since $$0\le {g(x)\over|x-a|}\le k|x-a|$$by squeeze theorem we have $$\lim_{x\to a^+} \frac{g(x) - g(a)}{x - a }=0$$For the left limit we obtain$$\lim_{x\to a^-} \frac{g(x) - g(a)}{x - a }=\lim_{x\to a^-} -\frac{g(x)}{|x - a| }$$while we have $$-k|x-a|\le -{g(x)\over x-a}\le 0$$then the Squeeze theorem is still applicable and we finally have $$\lim_{x\to a} \frac{g(x) - g(a)}{x - a }=0$$or $$g'(a)=0$$
A: Note that $\lim_{x\to a} k^2|x-a|=0$. So you get that $|\frac{g(x)-g(a)}{x-a}|$ is non negative on the one hand, and on the other hand it is bounded from above by a function that converges to $0$ when $x\to a$. Hence by the squeeze theorem the limit exists and equals $0$. Of course we also use the known fact that a function converges to $0$ if and only if its absolute value converges to $0$. 
