L'Hopital's Rule on Limits with Trigonometric Functions

Question

How do trigonometric functions affect the solving of limit as x grows arbitrarily large? Aren't trigonometric functions cyclical - meaning they wouldn't have any answer for $\lim_{x\to\infty}$?
How is the problem $\lim_{x\to \infty} (1-{\cos{(3x)}})^x$ solved?

Answer 1

This limit does not exist.

Take $x_n=\frac{\pi}{6}+2\pi n$ and $x_n=\frac{\pi}{12}+2\pi n$, where $n\rightarrow+\infty$ and $n\in\mathbb N$.

Answer 2

You could write this particular function as $$\exp[x\log(1-\cos(3x))],$$ and the problem simply reduces to taking the limit of $x\log(1-\cos(3x))$ since $\exp{}$ is continuous.

But as $x$ goes to $+\infty,$ the argument of the $\log{}$ only hurtles continuously between $1$ and $2$ because of the cosine, so that the logarithm bounces continuously between $0$ and $\log 2.$ The other factor, $x,$ in the meantime grows without bound.

So combining these two behaviours, you see that you have a function which gets arbitrarily large in magnitude but fluctuates continuously between positive and negative values. So the limit does not exist.

There is no need to use L'hopital here. One uses it in cases where one has a limiting value of the form $0/0,$ and cases that can be reduced to this. No such thing happens here. There simply isn't any limiting value.