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 -> 0) ( int_0^sqrtx cos(pi/2*e^(t^2))dt)/ (xsqrtx)`

That is `0/0` indeterminate so we use LHR and differentiate top and bottom wrt x.

For the top use the Liebnitz Integral Rule, which applies here as:

• `bb( d/dx ( int_0^(v(x)) f(s) \ ds ) = f(v(x)) * v^'(x) )`

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

`= lim_(x -> 0) ( cos(pi/2*e^(x)) )/ (3 x)`

That is still `0/0` indeterminate so we go again with LHR.

`= lim_(x -> 0) ( pi/2 e^x * - sin(pi/2*e^(x)) )/ (3 ) = ((pi/2)(-sin (pi/2)) ) /3= - pi /6`