In this question the asker mentions that the limit of this does not exist: $$\lim_{x\to \infty} \frac{1}{1+\cos(x)}$$ Graphically I can see that the limit doesn't exist, but I'd like to know what the proof is. I'm also wondering if there is a general rule that can be applied to any limit to tell if the limit exists.

Sorry if this question is a bit dumb, and thanks in advance for any answers.

  • 3
    $\begingroup$ You could use the fact that the function is periodic and nonconstant. $\endgroup$ – Travis Willse Mar 24 '16 at 18:24
  • $\begingroup$ So will any periodic function not have a limit at infinity? $\endgroup$ – BombSite_A Mar 24 '16 at 18:28
  • $\begingroup$ Any nonconstant periodic function, as if $\lim_{x \to \infty} f(x) = a$, then $\lim_{k \to \infty} f(a_k) = a$ for any sequence $a_k$ such that $\lim_{k \to \infty} a_k = \infty$. $\endgroup$ – Travis Willse Mar 24 '16 at 18:30
  • $\begingroup$ Because there is no interval of type $(a,\infty)$ where the function is defined. For the limit $\lim_{x\to\infty}f(x)$ to exist a precondition is that $f(x)$ must be defined in some interval of type $(a,\infty)$. $\endgroup$ – Paramanand Singh Mar 25 '16 at 7:09

enter image description here

Let $f$ be a function defined on $\mathbb{R}$ and $\ell\in \mathbb{R}$. Then $\displaystyle \lim_{x \rightarrow +\infty}f(x)=\ell$ if and only if for every sequence $(x_n)$ that tends to $+\infty$, we get $f(x_n) \rightarrow \ell$.

So, in order to prove that a limit $\displaystyle \lim_{x \rightarrow +\infty}f(x)$ does not exist, it is enough to prove that there are sequences $(x_n),~(y_n)$ that tend to $+\infty$ and $\ell_1,~\ell_2\in \mathbb{R}$, such that $\displaystyle \lim_{n \rightarrow +\infty}f(x_n)=\ell_1\neq \ell_2=\displaystyle \lim_{n \rightarrow +\infty}f(y_n).$

In our case: take $x_n=2\pi n,~y_n=2\pi n+\pi/2.$


As $x$ tends to infinity, there are arbitrarily large $x$ for which $\cos(x) = 1$ and for which $\cos(x) = 0$.

In the former case, the ratio at hand is $1/2$; in the latter case, it is $1$.

If a sequence attains two different values infinitely often, then it cannot converge.

A similar phenomenon is afoot with the limit taken of this expression.

Separately: You probably want your denominator to be slightly different, say, $2 + \cos(x)$.

Right now, you run the risk of encountering $x$ for which $\cos(x) = -1$, which will make the denominator of that expression a rather unwelcome $0$.

  • $\begingroup$ "If a sequence attains two different values infinitely often" - I'm guessing that should be "function" not "sequence", and $f(x) = \frac{sin(\frac 1x)}x$ is a function which attains the values 0 and 1 infinitely often but whose limit is 0 as $x$ goes to infinity. $\endgroup$ – user253751 Mar 25 '16 at 2:19
  • $\begingroup$ @immibis No, I quite deliberately wrote sequence rather than function. $\endgroup$ – Benjamin Dickman Mar 25 '16 at 2:41

$\cos(x)$ is periodic. For all values of $x$, $\cos(x)$ will take a value in $[-1,1]$.

It's probably easiest shown by an example. Let's take the cosine of one billion (in radians). $$\cos(10^9) \approx 0.84$$ Then $\frac{1}{1+0.84} \approx 0.54$. Now let's go just a little bit further in $x$, by $\pi/2$: $$\cos(10^9 + \pi/2) \approx -0.54$$

So then $\frac{1}{1-0.54} \approx 2.17$. Do you see why this won't converge? Cosine of $x$ oscillates forever. Consequently, the limit at infinity does not exist.


Suppose $\lim\limits_{x\to\infty} \dfrac 1 {1+\cos x} = L$.

Then for some number $x_0$, whenever $x>x_0$ then $ \dfrac 1 {1+\cos x}$ is between $L\pm 0.0001$.

There are some values of $x>x_0$ for which $\dfrac 1 {1+\cos x} = \dfrac 1 {1+0} = 1$ and there are some values of $x>x_0$ for which $\dfrac 1 {1+\cos x} = \dfrac 1 {1+1} = \dfrac 1 2$.

Therefore the numbers $1$ and $1/2$ are both between $L\pm0.0001$. That implies $1$ differs from $1/2$ by less than $2\times0.0001$.


If it is Known that any périodic function which admits a limit at + $\infty$ is necessarily constant then the résult follows immediatly because our function is périodic and not constant. The proof of this résult may be conducted by taking two séquences such that f takes a different constant value on each of which. (As Nikolaos skout did)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.