The value of #lim_(h→0)1/hint_2^(2+h)sin(x^2)dx#is?

The value of #lim_(h→0)1/hint_2^(2+h)sin(x^2)dx#is?

1 Answer
Jun 29, 2018

#lim_(h→0)(int_2^(2+h)sin(x^2)dx)/h#

This is #0/0# indeterminate so DLH's rule applies:

#= lim_(h→0)(d/(dh) int_2^(2+h)sin(x^2)dx)/(d/(dh)(h)#

Differentiating numerator using Liebnitz Rule:

#= lim_(h→0) sin(2+h)^2 *d/(dh)(2 + h)= sin 4#