# Differentiability at 0

Let $\alpha >1$. Show that if $|f(x)| \leq |x|^\alpha$, then $f$ is differentiable at $0$.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. –  Did Oct 17 '12 at 13:47
What I have at my disposition is the definition of a derivative as a limit. –  Carpediem Oct 17 '12 at 13:50
And what did you try to use it? –  Did Oct 17 '12 at 13:53
I used it as was advised below, but I don't know how to conclude the differentiability in 0. I know we have an expression similar to the limit but I don't know how to continue –  Carpediem Oct 17 '12 at 13:59
What you have is not an expression similar to the limit but a precise definition involving a quite definite limit. –  Did Oct 17 '12 at 15:21

Use the definition of the derivative. It is clear that $f(0)=0$. Note that if $h\ne 0$ then $$\left|\frac{f(h)-0}{h}\right| \le |h|^{\alpha-1}.$$ Since $\alpha\gt 1$, $|h|^{\alpha-1}\to 0$ as $h\to 0$.