lim_(x->1^(+)) (int_1^x cos(theta) d theta ) / (x-1) Evaluate the limit, using L’Hôpital’s Rule if necessary.

 
Givne to solve,
lim_(x->1^(+)) (int_1^x cos(theta) d theta ) / (x-1)
=lim_(x->1^(+)) ([sin(theta)]_1^x) / (x-1)
=lim_(x->1^(+)) ([sin(x)-sin(1)]) / (x-1)
when x-> 1+ then ([sin(x)-sin(1)]) / (x-1) = 0/0 form
so upon applying the L 'Hopital rule we get the solution as follows,
as for the general equation it is as follows
lim_(x->a) f(x)/g(x) is = 0/0 or (+-oo)/(+-oo) then by using the L'Hopital Rule we get  the solution with the  below form.
lim_(x->a) (f'(x))/(g'(x))
 
so , now evaluating
lim_(x->1^(+)) ([sin(x)-sin(1)]) / (x-1)
=lim_(x->1^(+)) (([sin(x)-sin(1)])') / ((x-1)')
=lim_(x->1^(+)) (cos(x)) / (1)
=lim_(x->1^(+)) (cos(x))
on plugging the value x= 1
we get
lim_(x->1^(+)) (cos(x))
=cos(1)