Why can't we determine the limit of $\cos x$ and $\sin x$ at $x=\infty $ or $x=-\infty$? I'm confused about why we can't determine the limit of $\cos x$ and $\sin x$ as  $x \to \infty$, even though they are defined over  $\mathbb{R}.$
When I use Wolfram Alpha, I get the following result (link to page):



which shows only that there are $2$ limits :$-1$ and $ 1 $.
Can someone show me why we can't determine $\lim \sin x$ and $\lim \cos x$ at  $x=\infty $ or $x=-\infty$ ?
Thank you for your help.
 A: Those two functions are periodic and their range is $[-1, 1]$ over each of the periods.  So you can find a sequence of $x_n\to\infty$ so that $\sin(x_n)$ converges to any value in $[-1, 1]$.  Limits only exist if all sequences converge to the same value.
A: They can't have a limit because they're periodic functions. What Wolfram Alpha outputs are the limit inferior and the limit superior of these functions, which always exist as soon as the functions are bounded. In case you haven't
Similarly seen these notions yet, by definition:
$$\limsup_{x\to\infty}f(x)=\lim_{x\to\infty}\sup_{t\ge x}f(t)$$
Note that the function $g(x)=\sup_{t\ge x}f(t)$ is non-increasing, hence it has a limit if $f$  is bounded from below.
Similarly,
$$\liminf_{x\to\infty}f(x)=\lim_{x\to\infty}\inf_{t\ge x}f(t)$$
which exists because the function $h(x)=\inf_{t\ge x}f(t)$ is non-decreasing, hence it has a limit if $f$  is bounded from above.
Also, one can prove that for any number $a\in[-1,1]$ there exists a sequence $(x_1, \dots, x_n,\dots) \to \infty$ such that $(f(x_1),\dots,f(x_n),\dots)\to a$.
A: It does not have two limits. What it says is that it cos or sin is always between -1 and 1 as x tends to infinity. Breaking news...
Actually it has no limit because by definition of the limit of a function f at +infinity, at a certain point A, for every x>=A, f(x) must stay "near" a certain value, and grow nearer and nearer as x increases. It is clearly not the case here, since cos and sin are oscillating continuously.
A: That's because sin and cos are cyclic functions, they do not have either a horizontal nor a vertical asymptote, they keep variating from y=1 to y=-1 all the way.
And easy way to think is picturing a trig circle and realising that as we keep increasing the angle, the value of sin or cosine varies in a closed interval of values (-1,1) 
A: Yes. Suppose for contradiction that $L = \lim_{x \to + \infty} \sin (x)$, where we may as well assume $-1 \leq L \leq 1$. Then for every $\epsilon > 0$, there exists $x_{0} > 0 $ such that if $x > x_{0}$, then $| \sin (x) - L | < \epsilon$.
Fix $0 < \epsilon < 1/2$.
Then either $\underline{y} = L - 2 \epsilon$, or $\overline{y} = L + 2 \epsilon$ is between $-1$ and $1$ (perhaps both); suppose $\underline{y}$ meets this assumption, and let $\theta = \arcsin (y)$, and pick $N \in \mathbb{N}$ big enough that $2 \pi N + \theta > x_{0}$. Then $| \sin ( 2 \pi N + \theta ) - L | = 2 \epsilon > \epsilon$, a contradiction.
A similar idea works for going to $- \infty$ and/or $\cos$.
