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Suppose that $f$ is differentiable at 0, and that $f(0) = 0$. Prove that $f(x) = xg(x)$ for some function $g$ which is continuous at 0.

This is a problem from Spivak's Calculus, namely problem 27 of Chapter 10. (This is not homework, but rather self-study.) I am not sure how to go about this proof. The hint given in the text is to consider that $g(x)$ can be written as $f(x)/x$, but this puzzles me, because then continuity of $g$ at 0 says that $\lim_{x \to 0} g(x) = g(0) = f(0)/0 = 0/0$.

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    Oh, I think I see where is the confusion. You can't define $g(0)$ using the formula $f(0)/0$ but you can use the fact that $f$ is differentiable at $0$ to define $g(0)$. Please consider carefully Moron's answer. It has more information than meets the eye ;)2010-12-30

2 Answers 2

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What is $\displaystyle \lim_{x \to 0} \frac{f(x) - f(0)}{x-0}$ ?

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    Adriá$n$: Alright, great! Tha$n$ks for $y$our help.2010-12-30
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Show that f'(0) = g(0) from the the continuity of $g$. Note that f'(0) = \lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h}.