# Limit of a trigonometric function

ehild
Homework Helper
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
Staff Emeritus
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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
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2019 Award
All extra-smartness aside, I pragmatically stick to the approach in #45 or this kind of exercises

SammyS
Staff Emeritus
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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.)

BvU
Gold Member