How do you find the Limit of [(sin x) * (sin^2 x)] / [1 -( cos x)] as x approaches 0?

1 Answer
Jun 20, 2016

Perform some conjugate multiplication and simplify to get lim_(x->0)(sinx*sin^2x)/(1-cosx)=0

Explanation:

Direct substitution produces indeterminate form 0/0, so we'll have to try something else.

Try multiplying (sinx*sin^2x)/(1-cosx) by (1+cosx)/(1+cosx):
(sinx*sin^2x)/(1-cosx)*(1+cosx)/(1+cosx)

=(sinx*sin^2x(1+cosx))/((1-cosx)(1+cosx))

=(sinx*sin^2x(1+cosx))/(1-cos^2x)

This technique is known as conjugate multiplication, and it works nearly every time. The idea is to use the difference of squares property (a-b)(a+b)=a^2-b^2 to simplify either the numerator or denominator (in this case the denominator).

Recall that sin^2x+cos^2x=1, or sin^2x=1-cos^2x. We can therefore replace the denominator, which is 1-cos^2x, with sin^2x:
((sinx)(sin^2x)(1+cosx))/(sin^2x)

Now the sin^2x cancels:
((sinx)(cancel(sin^2x))(1+cosx))/(cancel(sin^2x))
=(sinx)(1+cosx)

Finish by taking the limit of this expression:
lim_(x->0)(sinx)(1+cosx)
=lim_(x->0)(sinx)lim_(x->0)(1+cosx)
=(0)(2)
=0