# Limit of an integral where (in my opinion) it is not possible to apply de l'Hopital rule

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?

• 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 Apr 20, 2021 at 14:44
• Ok thanks for confirming my ideas! Apr 20, 2021 at 15:01