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`lim_(x->1) (arctanx-pi/4)/(x-1)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

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`

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