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I know that when $f(0)=0$,the integral $$I=\int_0^\varepsilon \frac{1}{f(x)}dx $$ may not diverge (like $f=\sqrt{x}$.

So I want to know if a function $f$ is differentiable in a neighborhood of $0$ such that $f(0)=f’(0)=0$, can we say that for any $\varepsilon$, the integral $I$ must diverge.

I don’t know how to prove it ,or give a counterexample.

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  • $\begingroup$ What's the relationship between $f(x)$ and $I$? $\endgroup$
    – Andrei
    Oct 26 at 14:41
  • $\begingroup$ @5xum sorry! I have rewritten the equation. $\endgroup$
    – zik2019
    Oct 26 at 14:42
  • $\begingroup$ @Andrei Sorry! I have rewritten the equation. $\endgroup$
    – zik2019
    Oct 26 at 14:43
  • $\begingroup$ Can you assume that $f'$ is continuous on some interval $(0,t)$ or do you specifically want to avoid that? $\endgroup$
    – ajr
    Oct 26 at 14:48
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Your conditions are sufficient to prove that $$\lim_{x\to 0}\frac{f(x)}{x} = 0$$


Method 1:

If $f$ is differentiable at $x_0$, then, by the definition of differentiability, there exists some function $o$ such that, for small values of $h$, you have

$$f(x_0+h)=f(x_0)+f'(x_0)\cdot h + o(h)$$

and the limit $$\lim_{h\to 0}\frac{o(h)}{h}=0.$$

In your case, $x_0=0$ gives you

$$f(x)=0+0\cdot x + o(x)=o(x)$$

so you basically know that

$$\lim_{x\to 0}\frac{f(x)}{x} = 0$$


Method 2:

By another definition of $f'$, you have

$$0=f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(h)}{h} = \lim_{h\to0}\frac{f(h)-0}{h} = \lim_{h\to0}\frac{f(h)}{h}$$


In both cases, once you have the limit, it is trivial to show that for values of $x$ close to $0$, you have $|f(x)|<|x|$, or, for positive values of $x$, simply $|f(x)|<x$.

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