
differentiability proof
a)
Suppose f(x)=xg(x) for some function g which is continuous at 0. Prove that f is differentiable at 0 and find f'(0) in terms of g.
b)
Suppose 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. Hint: what happens if you try to write g(x)=f(x)/x ?

1) $\displaystyle f(h) = h g(h) = h(g(0) + o(1)) = f(0) + hg(0) + o(h)$ so rearranging gives $\displaystyle {f(h)  f(0)\over h} = g(0) + o(1)$, or sending $\displaystyle h\to 0$, $\displaystyle f'(0) = g(0)$.
2) It comes down to showing that the limiting ratio $\displaystyle f(x)\over x$ exists. Use l'Hopitals rule. Note that the result is in agreement with (1), which is always good!