Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks in advance.

share|improve this question
16  
In the immortal words of Lindsay Lohan... –  Qiaochu Yuan Jan 19 '11 at 15:21
2  
@Qiaochu: your joke eludes me. Citation please? –  Willie Wong Jan 19 '11 at 19:18
4  
@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
2  
thank you Qiaochu! Damn that was worth wasting 9 minutes of my life! –  Arjang Jan 21 '11 at 8:50

4 Answers 4

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

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

share|improve this answer

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

share|improve this answer

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 $|f(x)-f(y)|=\left|\limsup_{x \to \infty}\cos x - \liminf_{n \to \infty} \cos x\right|\geq\varepsilon>0$.

share|improve this answer

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 -∞.

share|improve this answer
    
Your notation is misleading. What are $n$ and $N$? Are they positive integers? –  Aryabhata Jan 19 '11 at 13:14
4  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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