1. ## limit of integral

Evaluate attachment

(I uploaded the equation rather than typing it out as my previous post).

Thanks.

2. Originally Posted by arianas89
Evaluate attachment

(I uploaded the equation rather than typing it out as my previous post).

Thanks.
$\lim_{x\to{0}}\frac{1}{x^3}\int_0^x \sin(t^3)\,dt=\lim_{x\to{0}}\frac{\int_0^x \sin(t^3)\,dt}{x^3}$

Applying L'Hopital's Rule:

$\lim_{x\to{0}}\frac{\int_0^x \sin(t^3)\,dt}{x^3}=\lim_{x\to{0}}\frac{\sin(x^3)} {3x^2}$

Apply L'Hopital's Rule again...

$\lim_{x\to{0}}\frac{\sin(x^3)}{3x^2}=\lim_{x\to{0} }\frac{3x^2\cos(x^3)}{6x}=\lim_{x\to{0}}\frac{1}{2 }x\cos(x^3)=\color{red}\boxed{0}$

I hope this clarifies things!

--Chris

P.S. Anyone who reads this, is there another way to do this w/o L'Hopital's Rule?? (ahemm....Mathstud...)

3. Hello,
P.S. Anyone who reads this, is there another way to do this w/o L'Hopital's Rule??
I've never done it this way, so it's experimental ^^

When x approaches 0, one can say that t and hence t^3 approaches 0 because it's restricted below by 0 and above by x.

So $\sin(t^3) \sim t^3$

Therefore $\lim_{x \to 0} \frac{\int_0^x \sin(t^3) \ dt}{x^3}=\lim_{x \to 0} \frac{\int_0^x t^3 \ dt}{x^3}=\lim_{x \to 0} \frac{x^4}{4x^3}=\lim_{x \to 0} \frac x4=0$

4. Originally Posted by Moo
Hello,

I've never done it this way, so it's experimental ^^

When x approaches 0, one can say that t and hence t^3 approaches 0 because it's restricted below by 0 and above by x.

So $\sin(t^3) \sim t^3$

Therefore $\lim_{x \to 0} \frac{\int_0^x \sin(t^3) \ dt}{x^3}=\lim_{x \to 0} \frac{\int_0^x t^3 \ dt}{x^3}=\lim_{x \to 0} \frac{x^4}{4x^3}=\lim_{x \to 0} \frac x4=0$
(OP) you can do this for as you noted we are considering $\sin(x^3)$ in the subset of the reals described by the limits of integration and the radius of this region centered at zero is approaching zero. Therefore you
may apply asymptotic equivalences.