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...

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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)`