# `lim_(x->0)arcsinx/x` Evaluate the limit, using L’Hôpital’s Rule if necessary. Given to solve,

`lim_(x->0)arcsinx/x`

as `x-> 0` we get `arcsinx/x = 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...

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Given to solve,

`lim_(x->0)arcsinx/x`

as `x-> 0` we get `arcsinx/x = 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)arcsinx/x`

=`lim_(x->0)((arcsinx)')/((x)')`

as we know that `(arcsinx)' = 1/(sqrt(1-x^2))`

so,

=`lim_(x->0)(1/(sqrt(1-x^2)))/(1)`

=`lim_(x->0)(1/(sqrt(1-x^2)))`

upon plugging the value` x=0` ,

=` (1/(sqrt(1-(0)^2)))`

`= 1/sqrt(1)`

= `1`

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