What is the limit as $x\to\infty$ of $\cos x$?
Thanks in advance.
What is the limit as $x\to\infty$ of $\cos x$?
Thanks in advance.
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
$\lim \sup_{x\to\infty} \cos(x) = 1$ $\lim \inf_{x\to\infty} \cos(x) = -1$
The limit does not exist because $\cos{(2\pi n)} = 1$ for $n \in \mathbb{Z}$ and $\cos{(\pi + 2 \pi n)} = -1$ for $n \in \mathbb{Z}$.
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 $|f(x)-f(y)|=\left|\limsup_{x \to \infty}\cos x - \liminf_{n \to \infty} \cos x\right|\geq\varepsilon>0$.
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|
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 -∞
.