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

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Answer 1 · 2016-10-24T13:10:42

The limit is #2#.

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#