Question

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

Answer 1

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`