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...

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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)`