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I am having a problem with this exercise. Please help.

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.2012-10-17
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    What I have at my disposition is the definition of a derivative as a limit.2012-10-17
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    And what did you try to use it?2012-10-17
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    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 continue2012-10-17
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    What you have is not *an expression similar to the limit* but a precise definition involving a quite definite limit.2012-10-17

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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 $h\to 0$.