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 .

  • 4
    $\begingroup$ 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? $\endgroup$ Aug 8, 2015 at 22:50
  • $\begingroup$ Also using your note, $z\to\infty$ means $x\to 0$, so that approach would not work. $\endgroup$ Aug 8, 2015 at 22:55
  • $\begingroup$ this is the problem in this limit $\endgroup$ Aug 8, 2015 at 22:55
  • $\begingroup$ I said also it is not work , only i see may there is a relationship bewteen 2 $\endgroup$ Aug 8, 2015 at 22:56
  • 1
    $\begingroup$ @user3002473 Constant functions are periodic but their limit as $x\to\infty$ exists. One must also show that the function is not constant. $\endgroup$
    – Guest
    Aug 8, 2015 at 23:19

1 Answer 1


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.

  • $\begingroup$ sorry to tel you that x go to infty $\endgroup$ 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. $\endgroup$ 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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .