is this function infinitely differentiable at $x=0$?

Question

Let $f(x)$ be an infinitely differentiable function defined on real line $R$. Assuming $f(0)=0$, define $g(x)=f(x)/x$ if $x$ is not zero, and $g(0)=f'(0)$. My question is whether $g$ is infinitely differetiable? A further question is that if $f$ is assumed analytic on real line $R$, and $g$ is defined as above, then would $g$ be analytic?

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It is evident that $g$ is continous since $$lim_{x\rightarrow 0}(f(x)-f(0))/(x-0)=f'(0)=g(0)$$.

Answer 1

Write $$ f(x) = x f'(0) + \int^x_0 (x- u) f''(u)\, du.$$ (Check this by differentiating - but the integral is 'the' (an?) integral form of the remainder of the T.S.) Make the substitution $u = t x$... to conclude that your $g$ is infinitely differentiable.

And yes, $g$ is analytic if $f$ is: write the power series out...

Answer 2

no, think about the case of $f(x)=e^x-1$ if $f^{(n)}(0)\neq0$, then $g^{(n)}(x)$can't be differentiated at $x\neq0$