If f is differentiable at a, then f is Lipschitz of order 1 at a.
[attempt]
lim_x->a {(f(x)-f(a))/(x-a)} = f'(a).
hmm if I can somehow change this to f'(a)(x-a)=f(x)-f(a). And let C be a constant such that C > f'(a) then I would be done...
Any help?
Side-Note: Last time I asked something about Lipschitz at a point, I was met with comments asking what this meant, so I'm assuming this isn't standard usage? Anyways what I meant by "f is Lipschitz of order 1 at a": f is Lipschitz continuous at a if there exists a neighborhood of a such that |f(y)−f(x)|< C|y−x|.
EDIT: I have understood that lim_x->a {(f(x)-f(a))/(x-a)} = f'(a) can be changed to f'(a)(x-a)=f(x)-f(a) in a nhbd of a. And we're looking for Lipschitz at point a, so we only consider the nbhd of a anyways. So I think that should complete the proof (coupled with user15464's idea. thanks for the help!