6

Given a function $f:\mathbb{R}\to\mathbb{R}$ twice-derivable at $x=0$ with $f(0)=0$.

Define $g(x):=\frac{f(x)}{x}$ for all $x\neq 0$ and $g(0)=f'(0)$. Clearly, $g$ is continuous since $$ \lim_{x\to 0} g(x) =\lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \frac{f(x) - f(0)}{x-0} = f'(0). $$

Solving a problem, I got the limit $\lim_{x\to 0} \frac{g(x)}{x} =\lim_{x\to 0}\frac{f(x)}{x^2} $, which I can't manage. I suppose that it exists and is related with $f''(0)$. Could anyone show how to compute $\lim_{x\to 0} \frac{g(x)}{x}$? Thanks in advance.

EDIT: In a solution-book I read $g'(0)=f''(0) /2$. Can this be typo? Maybe they forgot some assumptions in the problem.

Senna
  • 1,233
  • 1
    $f(x)=x$ and $f(x)=x^2$ both satisfy the conditions and they give different answers. – Gary May 09 '21 at 09:08
  • 4
    I find it a bit strange that this question got 5 upvotes in less than 3 minutes of posting – tryst with freedom May 09 '21 at 09:08
  • 3
    @Buraian I upvoted since it is "well-asked". – VIVID May 09 '21 at 09:09
  • Not so "well-asked": I doubt you really meant to ask for $\lim_{x\to 0} \frac{g(x)}{x}$ (which obviously does not exist if $g(0)=f'(0)\ne0$; the two trivial answers below should rather be comments about that). As for the more reasonable question of your edit, look at this duplicate: https://math.stackexchange.com/questions/2998034 – Anne Bauval Jun 20 '23 at 13:28

2 Answers2

4

The example $f(x)=x$ shows that the limit need not exist.

Kavi Rama Murthy
  • 311,013
1

Note that the limit $$\lim_{x\to0} \frac{g(x)}{x} = \lim_{x\to0} \frac{f(x)}{x^2} = \lim_{x\to0}\frac{f'(x)}{2x} $$ may exist iff $\lim_{x\to0}f'(x) = 0$.

VIVID
  • 11,604
  • That is what I guess... However, in my book it appears $ g'(0) = f''(0) / 2 $... but there is no assumption $f'(0)=0$. – Senna May 09 '21 at 09:21
  • 1
    @Senna Indeed, if $f'(x\to 0) = 0$ then, L'hospital gives that result. Maybe your book forgot about it or just a typo – VIVID May 09 '21 at 10:30
  • $\lim_{x\to0} \frac{f(x)}{x^2} = \lim_{x\to0}\frac{f'(x)}{2x}$ (L'Hospital) cannot be used if you do not know yet that $\lim_0f'=0.$ Anyway, directly: $\lim_{x\to0}\frac{g(x)}x$ does not exist if $\lim_0g\ne0.$ – Anne Bauval Jun 20 '23 at 13:35