How do you evaluate the limit #(sin^2xcosx)/(1-cosx)# as x approaches #0#?

1 Answer
Oct 24, 2016

The limit is #2#.

Explanation:

Let's start by simplifying the function because if we substitute directly we get an indeterminate form.

#lim_(x ->0)(sin^2xcosx)/(1 - cosx)#

We will apply the identity #sin^2x+ cos^2x = 1 -> sin^2x = 1 - cos^2x#.

#=lim_(x ->0)((1 - cos^2x)cosx)/(1 - cosx)#

We can factor #1 - cos^2x# as a difference of squares

#=lim_(x->0) ((1 + cosx)(1 - cosx)(cosx))/(1 - cosx)#

#=lim_(x-> 0) ((1 + cosx)cancel(1 - cosx)(cosx))/(cancel(1- cosx))#

#=lim_(x-> 0) cosx +cos^2x#

#=cos^2(0) + cos(0)#

#=1 + 1#

#= 2#