Question

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

Answer 1

`lim_(x->0)(sinxsin2x)/(1-cosx) = 4`

Explanation:

At the limit, fraction is of indeterminate form `0/0` so we can use L'Hopital's rule.

`lim_(x->0) (sinxsin2x)/(1-cosx) = lim_(x->0) (d/(dx)(sinxsin2x))/(d/(dx)(1-cosx))`

`=lim_(x->0)(cosxsin2x + 2sinxcos2x)/(sinx)`

Still of indeterminate form so apply L'Hopital's again:

`=lim_(x->0)(d/(dx)(cosxsin2x + 2sinxcos2x))/(d/(dx)(sinx))`

`lim_(x->0)(-sinxsin2x + 2cosxcos2x + 2cosxcos2x - 4sinxsin2x)/(cosx)`

`lim_(x->0)(4cosxcos2x - 5sinxsin2x)/(cosx)`

As `sin(0) = 0 and cos(0) = 1`

`=(4-0)/1 = 4`

`lim_(x->0)(sinxsin2x)/(1-cosx) = 4`

Answer 2

See below for an alternative approach using trig identities.

Explanation:

Euan's method of using L'Hopital's Rule is perfectly valid, but if you have not been exposed to it yet there is another way to solve the problem.

Recall the double angle identity for sine:
`sin2x=2sinxcosx`

Using this identity, we can rewrite the limit as follows:
`lim_(x->0)((sinx)(sin2x))/(1-cosx)`

`->lim_(x->0)((sinx)(2sinxcosx))/(1-cosx)`

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

Also recall the Pythagorean Identity `sin^2x+cos^2x=1`. Subtracting `cos^2x` from both sides, we obtain `sin^2x=1-cos^2x`. We can therefore rewrite the limit further:
`lim_(x->0)(2sin^2xcosx)/(1-cosx)`

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

Note that, by the difference of squares property, `1-cos^2x=(1)^2-(cosx)^2=(1-cosx)(1+cosx)`, since:
`a^2-b^2=(a-b)(a+b)`

Applying this to the problem, we have:
`lim_(x->0)(2(1-cos^2x)cosx)/(1-cosx)`

`->lim_(x->0)(2(1-cosx)(1+cosx)cosx)/(1-cosx)`

`1-cosx` cancels:
`lim_(x->0)(2cancel((1-cosx))(1+cosx)cosx)/cancel((1-cosx))`

`->lim_(x->0)2(1+cosx)cosx`

We can now attempt a direct substitution:
`lim_(x->0)2(1+cosx)cosx`

`->2(1+cos(0))cos(0)`

`->2(1+1)(1)`

`=2(2)(1)=4`

As expected, the result is the same.