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What is the limit as $x\to\infty$ of $\cos x$?

Thanks in advance.

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In the immortal words of Lindsay Lohan... –  Qiaochu Yuan Jan 19 '11 at 15:21
@Qiaochu: your joke eludes me. Citation please? –  Willie Wong Jan 19 '11 at 19:18
@Willie: youtube.com/watch?v=QIMSC-RWvF8 (skip to 7:40 for the problem and 8:10 or so for its solution). –  Qiaochu Yuan Jan 19 '11 at 19:22
thank you Qiaochu! Damn that was worth wasting 9 minutes of my life! –  Arjang Jan 21 '11 at 8:50
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4 Answers

The limit does not exist. It oscillates between -1 and 1. Just so that you know, the limit supremum or infimum as $x \to \infty$ is given as

$$\lim \sup_{x\to\infty} \cos(x) = 1$$ $$\lim \inf_{x\to\infty} \cos(x) = -1$$

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The limit does not exist because $\cos{(2\pi n)} = 1$ for $n \in \mathbb{Z}$ and $\cos{(\pi + 2 \pi n)} = -1$ for $n \in \mathbb{Z}$.

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There is no limit, $\lim_{x \to \infty} \cos x$, since $\cos$ oscillates between -1 and 1.

A bit more detailed: We say that a function $f(x)$ has a limit as $x \to \infty$ if there exists a real number $a$ (called the limit) such that $|f(x)-a|$ can be made arbitrarily small for all $x$ which are "large enough". "Large enough" and arbitrarily small means that for all $\varepsilon > 0$, we should be able to find a number $N$ such that $|f(x)-a| < \varepsilon$ for all $x > N$.

In the case of $f(x) = \cos x$ we can't do this when $\varepsilon$ is small. Independent of which $a$ we are trying out, we can always find a large enough $x$ such that $|\cos x - a| > \frac{1}{2}$ (for example), and infinitely many of them (since $\cos$ is periodic).

As Roupam Ghosh stated in his answer, the $\limsup$ and $\liminf$ for $\cos x$ as $x \to \infty$ are not equal. This gives that the limit does not exist, since then you can always find sufficiently large $x,y$ such that $|f(x)-f(y)|=\left|\limsup_{x \to \infty}\cos x - \liminf_{n \to \infty} \cos x\right|\geq\varepsilon>0$.

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I believe that the proofs given are not rigorous enough, and are not necessarily easy to understand to someone who is not familiar with the material, so I'll try a more rigorous approach. I'll assume every variable I'm referring to is a real number in R.

Assume that there is a limit k. In that case it must be that for every d there exist N>0 so that if n>N we have that |cos(n)-k|<d.

Since cos(x+n2pi)=cos(x), it is easy to see that for every N we can find a number n0>N for which cos(n0)=0, and another number n1 for which cos(n0)=1. Try to prove that, it's not so hard.

Let d=1/2, whatever k would be, we'll have that either |cos(n0)-k|=|k|>d=1/2 or |cos(n1)-k|=|1-k|>d=1/2.

Therefor no number is the limit of limn→∞cos(n). Since cos(n) is bounded it cannot be that the limit is or -∞.

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Your notation is misleading. What are $n$ and $N$? Are they positive integers? –  Aryabhata Jan 19 '11 at 13:14
I don't quite see your point. If $a = \lim_{x \to \infty} \cos{(x)}$ existed then by definition it would be the limit of every sequence of real numbers $x_{n} \to \infty$, so it suffices to exhibit two sequences converging to different limits. –  t.b. Jan 19 '11 at 14:26
Good point! I think integer or real doesn't really matter. Tried to clarify this issue. –  Elazar Leibovich Jan 19 '11 at 14:48
@Theo, my point is, (1) you have to spell out the reason like you did, and I didn't find that in the answers, (2) sometimes the TA wants the answer in an epsilon-delta form. –  Elazar Leibovich Jan 19 '11 at 14:50
@Elazar: We are not being graded by a TA on this site. You can supply a lot of detail if you like, but it is not expected that full details are spelled out in each answer, especially if the question is for homework. –  Jonas Meyer Jan 21 '11 at 7:29
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