lim_(x->0) sin(ax)/sin(bx) , a,b!=0 Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,
lim_(x->0) ((sin ax)/(sin bx))
as x->0 then the ((sin ax)/(sin bx)) =sin(0)/sin(0) =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) ((sin ax)/(sin bx))
=lim_(x->0) ((sin ax)')/((sin bx)')
=lim_(x->0) ((cos ax)(a))/((cos bx)(b))
upon plugging the value of x= 0 we get
=((cos a(0))(a))/((cos b(0))(b))
= (a/b) (cos 0/cos 0)
= (a/b) (1/1)
= (a/b)

Given to solve,
lim_(x->0) (sin(ax))/(sin(bx))
as we know that
as x-> 0 then sin(x) -> x
so we can state that
sin(ax) = ax and
sin(bx) = bx            since x-> 0
so ,
lim_(x->0) (sin(ax))/(sin(bx))
= lim_(x->0) (ax)/(bx)
= lim_(x->0) (a/b)
= (a/b)