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

Question

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

Thanks in advance.

Answer 1

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

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

Answer 2

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

Answer 3

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

Answer 4

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 lim_n→∞cos(n). Since cos(n) is bounded it cannot be that the limit is ∞ or -∞.