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

1 Answer
Ultrilliam · Anjali G
Jun 11, 2018

#- 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#

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