How do I evaluate this limit :$\displaystyle \lim_{x\to \infty} (1+\cos x)^\frac{1}{\cos x}$?

Question

I would like to know if this :$$ \lim_{x\to \infty} (1+\cos x)^\frac{1}{\cos x}$$ does exist and how do i evaluate it ?.

Note : I have tried to use the standard limit : $$ \lim_{z\to \infty} \left(1+\frac{1}{z}\right)^z=e$$ using $\cos x=1/z $ but i can't succeed

Thank you for any help .

Hint: Use the fact that $\cos$ is periodic to show that $(1 + \cos(x))^{\sec(x)}$ is periodic as well. Then, if a function $f(x)$ is periodic, does $\lim_{x\to\infty}f(x)$ exist? — user3002473, Aug 8, 2015 at 22:50
Also using your note, $z\to\infty$ means $x\to 0$, so that approach would not work. — John Molokach, Aug 8, 2015 at 22:55
I said also it is not work , only i see may there is a relationship bewteen 2 — zeraoulia rafik, Aug 8, 2015 at 22:56
@user3002473 Constant functions are periodic but their limit as $x\to\infty$ exists. One must also show that the function is not constant. — Guest, Aug 8, 2015 at 23:19

DanielWainfleet · Accepted Answer · 2015-08-08 23:13:44Z

6

For any natural number $n$, if $x=2n\pi$ ,the value of the function is 2; if $x=(2n+1/2)\pi$, the value is 1. So no convergence.

answered Aug 8, 2015 at 23:13

DanielWainfleet

56.8k4 gold badges29 silver badges71 bronze badges

$\begingroup$ sorry to tel you that x go to infty $\endgroup$
– zeraoulia rafik
Aug 8, 2015 at 23:15
1

$\begingroup$ @zeraouliarafik Substitute $x = 2n\pi$ and let $n$ go to $\infty$. Then $x \to \infty$ as well. $\endgroup$
– user3002473
Aug 8, 2015 at 23:16
1

$\begingroup$ This is clever, so you can argue that such limits of periodic functions don't exist because they are not unique. +1 $\endgroup$
– Tucker
Aug 8, 2015 at 23:21

Add a comment |

1 Answer 1