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I want to prove that the following limit is $0$: $$\lim_{x\to 0}\frac{1}{x}\int_0^{x^3}f(t)\,dt$$ The only hypothesis of the problem are: $f$ is bounded and integrable.

So I have thought that since I don'thave the hypothessi of continuity of the function $f$ I can't say that surely $\int_0^{x^3}f(t)\,dt$ is differentiable, as it is stated from the fundamental theorem of integral calculus. So for this reason I can't apply for instance the de l'Hopital theorem.

But I know that the function is boundend so this means that: $\exists M>0\, s.t.\,$ $-M\leq f(x)\leq M$ $\forall x\in \mathbb{R}$. And so: $$-M\leq f(x)\leq M\implies -\int_0^{x^3}M\, dt\leq \int_0^{x^3} f(t)\, dt\leq \int_0^{x^2} M\, dt\implies$$ $$\implies 0=-\lim_{x\to 0}Mx^2=\lim_{x\to 0}-\frac{1}{x}\int_0^{x^3} M\, dt\leq\lim_{x\to 0}\frac{1}{x}\int_0^{x^3}f(t)\,dt\leq \lim_{x\to 0}\frac{1}{x}\int_0^{x^3} M\, dt=\lim_{x\to 0}Mx^2=0$$ So $\lim_{x\to 0}\frac{1}{x}\int_0^{x^3}f(t)\,dt=0$.

Do you think is my idea correct? And above all the remark on the fact that I can't apply the de l'Hopital is right?

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  • $\begingroup$ Yes, your idea is correct. Also the remark about de l’Hopital seems right to me, $\int_0^{x^3} f(x)dx$ can fail to be differentiable $\endgroup$ Apr 20, 2021 at 14:44
  • $\begingroup$ Ok thanks for confirming my ideas! $\endgroup$
    – pawel
    Apr 20, 2021 at 15:01

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