Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 20

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)

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