Given to solve,

`lim_(x->1) (arctanx-pi/4)/(x-1)`

as `x->1` then the `(arctanx-pi/4)/(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...

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Given to solve,

`lim_(x->1) (arctanx-pi/4)/(x-1)`

as `x->1` then the `(arctanx-pi/4)/(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) ((arctanx-pi/4)')/((x-1)')`

=`lim_(x->1) ((1/(1+x^2))-0)/(1)`

=`lim_(x->1) ((1/(1+x^2)))`

so , now plugging the value of the `x =1` then we get

=`((1/(1+(1)^2)))`

= `1/(1+1)`

=`1/2`