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...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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`