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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!

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    @MathMathCookie: Ah, good point.2011-11-08

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Choose an exponent $u$ and define a function $f$ by $f(0)=0$ and, for $x\ne0$, $ \color{red}{f(x)=x^u\cos(\pi x^{-1})}. $ Then $f$ is continuous at $0$ (and everywhere else) if $u>0$ and $f$ is differentiable at $0$ (and everywhere else) if $u>1$. But $x=(2n)^{-1}$ and $y=(2n+1)^{-1}$ for any positive integer $n$ yields $|x-y|=x-y=\Theta(n^{-2})$ while $\cos(\pi x^{-1})=+1$ and $\cos(\pi y^{-1})=-1$ hence $|f(x)-f(y)|=x^u+y^u=\Theta(n^{-u})$.

Finally, for every exponent $u$ in $(1,2)$, $f$ is differentiable everywhere and is not what you call Lipschitz continuous at $0$.