How do you find the limit #lim_(x->0)sin(x)/x# ?
1 Answer
Aug 18, 2014
We will use l'Hôpital's Rule.
l'Hôpital's rule states:
#lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))#
In this example,
Thus,
#lim_(x->0) (sinx)/x = lim_(x->0) (cosx)/(1)#
Quite clearly, this limit evaluates to