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`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)`

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