Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 40

Given to solve,
lim_(x->0) x/arctan(2x)
as x->0 then the x/arctan(2x) =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->0) x/arctan(2x)
=lim_(x->0) (x')/((arctan(2x))')
= lim_(x->0) 1/((arctan(2x))')

First let us compute the
(arctan(2x))'
let u= 2x
=>
so,
(arctan(2x))'= d/dx (arctan(2x))
=d/(du) (arctan(u)) d/dx (u) [as d/dx f(u) = d/(du) f(u) * d/dx (u)]
= (1/(u^2+1) d/dx (2x)
=(1/(u^2+1) )(2)
=(2/(u^2+1)) but u= 2x ,so
=2/((2x)^2+1)
= (2/(4(x)^2+1))

now coming back to the limits , we have
lim_(x->0) 1/(arctan(2x))'
= lim_(x->0) 1/(2/(4x^2+1))
as x->0 , we get
=1/((2/(4(0)^2+1)) )
= 1/2
so , we can state that ,
lim_(x->0) x/arctan(2x) = 1/2