Limit of a trigonometric function

[ehild]

Just need to get cos(3x) in terms of cos(x):

$\cos(3x)=4\cos^3(x)-3\cos(x)$​

The derivative of the numerator is then $\ 12\sin^3(x)\ .$

The derivative of the denominator then has a factor of $\ \sin^3(2x)\$ which is easily reduced.

$\sin^3(x)\$ in both numerator & denominator cancel. The answer is readily at hand.

Sammy is the greatest!

 

[SammyS]

As a followup to post #49, which showed that if you follow @ehild's advice, it is only necessary to apply L'Hôpital's rule one time:

This implies that simplifying OP's result (once corrected) after the first application of L'Hôpital's rule should give the desired result with no further need to appeal to L'Hôpital .

That is indeed the case.

 

[BvU]

All extra-smartness aside, I pragmatically stick to the approach in #45 or this kind of exercises

 

[SammyS]

All extra-smartness aside, I pragmatically stick to the approach in #45 or this kind of exercises

I agree that using the Taylor series approach can give the limit more quickly and with less chance for error. It's what I used to get my initial answer.

(Apparently I have some Dutch heritage.)

 

[chwala]

Good read am learning.