1. ## Integral

Hi,

I'm a bit stuck on this question:

Let g be differentiable in [-1,1]. g(0)=0 and g'(0)=1
Does the limit in the attached image exist (below)? If yes, calculate it. If no, explain why not.

I'm not really sure how to answer this at all. It came up on an exam I had recently.

Thanks

2. ## Re: Integral

First hint: Denote $h(x)=\int_{x^3}^{x^2}\left(\int_0^tg(s^2)ds\right) dt$ . Then $\lim_{x\to 0}h(x)=0$ , as a consequence

$\lim_{x\to 0}\frac{\sin \left(\int_{x^3}^{x^2}\left(\int_0^t g(s^2) ds\right) dt\right)}{x^8}=\lim_{x\to 0}\dfrac{h(x)}{x^8}$

P.S. Show some work and I will provide more hints if necessary.

3. ## Re: Integral

Originally Posted by FernandoRevilla
First hint: Denote $h(x)=\int_{x^3}^{x^2}\left(\int_0^tg(s^2)ds\right) dt$ . Then $\lim_{x\to 0}h(x)=0$ , as a consequence

$\lim_{x\to 0}\frac{\sin \left(\int_{x^3}^{x^2}\left(\int_0^t g(s^2) ds\right) dt\right)}{x^8}=\lim_{x\to 0}\dfrac{h(x)}{x^8}$

P.S. Show some work and I will provide more hints if necessary.
I don't know we got this: $\lim_{x\to 0}h(x)=0$.

Also did you mean the following on the last line?
$\lim_{x\to 0}\frac{\sin h(x)}{x^8}$

Does the limit then equal infinity according to L'Hopital?

Thanks for your help so far

4. ## Re: Integral

How do I know that the limit h(x) = 0 (when x approaches 0)?

5. ## Re: Integral

Originally Posted by durrrrrrrr
I don't know we got this: $\lim_{x\to 0}h(x)=0$.
$h$ is continuous in $[-1,1]$ and $h(0)=0$

Also did you mean the following on the last line? $\lim_{x\to 0}\frac{\sin h(x)}{x^8}$
We have $\lim_{x\to 0}\frac{\sin h(x)}{h(x)}=1$ so, $\lim_{x\to 0}\frac{\sin h(x)}{x^8}=\lim_{x\to 0}\frac{ h(x)}{x^8}$

Does the limit then equal infinity according to L'Hopital?
The limit presents the indetermination $0/0$ . Apply L'Hopital Rule.

6. ## Re: Integral

Originally Posted by durrrrrrrr
How do I know that the limit h(x) = 0 (when x approaches 0)?
Consider Taylor's theorem for $g(x)$ :

$g(x)=g(0)+xg'(0)+\rho(x)$

where $\lim_{x \to 0} \rho(x)/x =0$

CB