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, f(x) would be sinx, and g(x) would be x.

Thus,

lim_(x->0) (sinx)/x = lim_(x->0) (cosx)/(1)

Quite clearly, this limit evaluates to 1, since cos 0 is equal to 1.