Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 22

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