Question

How to find the limit for `lim x -> 0 ( 1/(xsqrtx) * int_0^sqrtx cos(pi/2*e^(t^2))dt)` with l'Hospital?

Answer 1

`- pi /6`

Explanation:

`lim_(x to 0) (int_0^sqrtx cos(pi/2*e^(t^2))dt)/(xsqrtx)`

This is `0/0` indeterminate so L'Hopital:

`= lim_(x to 0) (d/dx( int_0^sqrtx cos(pi/2*e^(t^2))dt))/(d/dx (x^(3/2) ))`

`= lim_(x to 0) ( cos(pi/2*e^((sqrtx)^2)) d/dx(sqrtx))/(3/2 x^(1/2) )`

`= lim_(x to 0) ( 1/2 cos(pi/2*e^(x)) )/(3/2 x ) = lim_(x to 0) ( cos(pi/2*e^(x)) )/(3 x )`

That is still `0/0` indeterminate.

`= lim_(x to 0) ( d/dx cos(pi/2*e^(x)) )/(d/dx 3 x )`

`= lim_(x to 0) (- sin (pi/2*e^(x))* pi/2 e^x )/(3 )`

With:

• `lim_(Q to 0) e^Q = 1`

`= (- sin (pi/2*1)* pi/2 (1) )/(3 ) = - pi /6`