How do you find the limit of #(sin²x)/(x)# as x approaches 0?

1 Answer
Aug 2, 2016

0

Explanation:

#lim_(x to 0) (sin²x)/(x)#

#lim_(x to 0) (sinx)/(x) * sin x#

#lim_(x to 0) (sinx)/(x) * lim_(x to 0) sin x#, we can seperate because sin x is continuous through the limit

# = 1 * 0 = 0 #

because #lim_(x to 0) (sinx)/(x) = 1# is a very well known limit

OR

#lim_(x to 0) (sin²x)/(x)#

is #0/0# indeterminate, so we can use L'Hopital

#= lim_(x to 0) (2 sinx cos x)/(1)#

#= lim_(x to 0) sin 2x = 0#