I have this Calculus 1 question as a homework and I was wondering how it is possible to prove that f'(0)=0 as the solution says.
I just can't seem to understand how to deal with this problem. Thank you everyone!
$\endgroup$ 54 Answers
$\begingroup$You don't need continuity of $f'$.
Suppose $d = f'(0) \neq 0$.
Then ${f(f(x)) \over f(x^2)} = { f(f(x)) - f(f(0)) \over f(x) - f(0)} { f(x) -f(0) \over x-0} {x-0\over x^2 -0} {x^2 - 0 \over f(x^2) -f(0)} = {1 \over x} { f(f(x)) - f(f(0)) \over f(x) - f(0)} { f(x) -f(0) \over x-0} {x^2 - 0 \over f(x^2) -f(0)}$.
Now note that $\lim_{x \to 0} { f(f(x)) - f(f(0)) \over f(x) - f(0)} { f(x) -f(0) \over x-0} {x^2 - 0 \over f(x^2) -f(0)} = d$.
$\endgroup$ 6 $\begingroup$Notice that $f(f(x))$ and $f(x^2)$ approach $0$ when $x$ goes to $0$ (by continuity of $f$).
Hence we can use l'Hôpital to obtain:
$\lim\limits_{x\to 0} \frac{f'(f(x))f'(x)}{2xf'(x^2)}=\pi$.
Suppose that $f'(0)=d$ with $d\neq 0$, then we can use continuity of $f'$ to obtain:
$\lim\limits_{x\to 0} \frac{f'(f(x))f'(x)}{f'(x^2)}=\frac{d^2}{d}=d$, and so $\lim\limits_{x\to 0} \frac{f'(f(x))f'(x)}{2xf'(x^2)}=\infty$.
We deduce $f'(0)=0$.
$\endgroup$ $\begingroup$f(x^2) is an even function. If f(x) was an odd function then f(f(+dx)) would have the opposite sign as f(f(-dx)). That would make it impossible for the limit as x -> 0+ and x ->0- of f(f(x))/f(x^2) to both be equal to pi. Thus f(x) must be locally even about x=0, and as an immediate consequence f'(x) must be zero at x=0.
$\endgroup$ $\begingroup$Taylor's theorem lets us write $$f(x)=f(0)+f'(0)\,x+o(|x|)=f'(0)\,x+o(|x|),$$ from where $$ \pi=\lim_{x\to0}\frac{f(f(x))}{f(x^2)}=\lim_{x\to0}\frac{f'(0)^2x+o(|x|)}{f'(0)\,x^2+o(|x|^2)}, $$ which is absurd if $f'(0)\neq0$.
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