lim_(x->0)arctanx/sinx Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,
lim_(x->0)arctanx/sinx
as x->0 on substituting we get
arctanx/sinx = 0/0
so by using 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)arctanx/sinx
=lim_(x->0)((arctanx)')/((sinx)')
= lim_(x->0)(1/(x^2 +1))/(cosx)
so now on applying x->0 ie x=0
=(1/(0^2 +1))/(cos0)
=1