What's the limit using L'Hopital's Rule

Lim x->0 1/(x^3) ∫sin(t^2) dt from 0 to x

∫sin(t^2) dt ' = sin(t^2) = sin(x^2)

sin(x^2) ' = 2x*cos(x^2)

2x*cos(x^2) ' = -4x^2*sin(x^2) + 2cos(x^2)

x^3 ' = 3x^2

3x^2 ' = 6x

6x ' = 6

lim x->0 ( -4x^2*sin(x^2) + 2cos(x^2))/6 = 1/3

1. take the derivative of upper and lowers separately to get [derivative of ∫sin(t^2) dt is simply sin(t^2)]

sin(t^2)/3x^2
2. take the derivatives again to get 2tcost(t^2)/6x
3. as t --> 0 , cos(t^2)=1 so you would have 2t/6x, therefore as t and x both approach 0 just simplify to get the final answer of 2/6 = !/3

1/3 is your final answer

good luck, I hope I explained it clearly

D(∫sin(t^2) dt from 0 to x) = sin(x^2)
D(x^3)=3x^2

Lim x->0 1/(x^3) ∫sin(t^2) dt from 0 to x =
= Lim x->0 sin(x^2)/(3x^2)=
= Lim x->0 (2x cos(x^2))/(6x) =

2x can be cancelled from both num and denominator

= Lim x->0 (cos(x^2))/3 = cos 0 /3 = 1/3