Evaluate attachment
(I uploaded the equation rather than typing it out as my previous post).
Thanks.
$\displaystyle \lim_{x\to{0}}\frac{1}{x^3}\int_0^x \sin(t^3)\,dt=\lim_{x\to{0}}\frac{\int_0^x \sin(t^3)\,dt}{x^3}$
Applying L'Hopital's Rule:
$\displaystyle \lim_{x\to{0}}\frac{\int_0^x \sin(t^3)\,dt}{x^3}=\lim_{x\to{0}}\frac{\sin(x^3)} {3x^2}$
Apply L'Hopital's Rule again...
$\displaystyle \lim_{x\to{0}}\frac{\sin(x^3)}{3x^2}=\lim_{x\to{0} }\frac{3x^2\cos(x^3)}{6x}=\lim_{x\to{0}}\frac{1}{2 }x\cos(x^3)=\color{red}\boxed{0}$
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...)
Hello,
I've never done it this way, so it's experimental ^^P.S. Anyone who reads this, is there another way to do this w/o L'Hopital's Rule??
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 $\displaystyle \sin(t^3) \sim t^3$
Therefore $\displaystyle \lim_{x \to 0} \frac{\int_0^x \sin(t^3) \ dt}{x^3}=\lim_{x \to 0} \frac{\int_0^x t^3 \ dt}{x^3}=\lim_{x \to 0} \frac{x^4}{4x^3}=\lim_{x \to 0} \frac x4=0$