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 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