limx>0 (1-cos2ax)/(1-cos2bx)

Question

limx->0 (1-cos2ax)/(1-cos2bx)

Answer 1

Limit as x→0 of (1 - cos2ax) / (1 - cos2bx) = ?

 

This is an indeterminate form, so use L'Hopital's rule.

 

(1 - cos2ax) / (1 - cos2bx) → (2a sin2ax) / (2b sin2bx)

 

This is another indeterminate form, so apply the rule again.

 

(2a sin2ax) / (2b sin2bx) → 4a²cos2ax / (4b²cos2bx)

 

Now let x→0 in both numerator and denominator.  You get 4a²/(4b²) = a²/b², and that is the limit.

 

It looks like you posted a few other problems involving indeterminate forms.  Usually, you solve them by applying L'Hopital's rule.  Sometimes, after applying the rule, you get a second indeterminate form.  You can try applying the rule again until you get a determinate form, as in this problem.

Answer 2

An attempt to evaluate this yields (1-1) divided by (1-1), which is indeterminate.

So use l'Hopital's rule.

 

The only problem is are those 2s supposed to be exponents, or are they part of the argument, i.e. cos(2ax)?

You will have to decide which, and that will affect the differentiations.