Sine of infinity? While considering limiting problems there are some situations when we have argument tending to infinity of sine or cosine function .
My book writes it as an "OSCILLATING number between $-1$ & $1$".
How is this possible?  
 A: What is oscilatting between $1$ and $-1$ is the sine (and the cosine). It follows from this that the limit cannot exist.
It's even worst with the tangent function: it keeps oscilatting between $-\infty$ and $+\infty$. The conclusion is the same, of course: $\lim_{x\to\pm\infty}\tan x$ does not exist.
A: "$\sin \infty$" (informal notation) is not a defined number because the function $\sin x$ is oscillating. For this reason,
$$\lim_{x\to\infty}\sin x$$ does not exist.
A: We know that $\sin x$ can takes any value that is inside $[-1,1]$
So, when we say x tends to infinity that means x is getting larger and larger. 
$$\lim_{x \rightarrow \infty}\sin x=DNE$$
Visual aid.

Can you see why the limit does not exist? It is a form of oscillatory series. It oscillates between $-1\leq x\leq 1$. $x\mapsto\infty$ It does not approach any fixed value of y.  
A: I have a slightly more creative and less conventional answer which I personally think is viable if you're not taking a calulus test.
On WolframAlpha if you do sin(infinity) you will get "-1 to 1." I am not sure how they got this answer but I definitely agree with it and here's why.
sin(∞) ?= [-1,1]
first of all, ∞=n+∞ for all n>0 and heres why:
x=n+x
using infinite iterations of recursive substitution we get
x=n+n+n+n+n+⋯
we can rewrite this as
∞
∑ n
i=1
we assume the above sum is equal to ∞ for all n>0, therefore x=∞
therefore ∞=n+∞ for all n>0
heres where we prove that sin(∞)=[-1,1]
if sin(∞)=sin(0)
then sin(∞+90)=sin(0+90)
but since ∞=90+∞,
sin(∞)=sin(0) and sin(90) with the initial assumption sin(∞)=sin(0)
we can extend this to all positive numbers replacing 90 with n
we thus can prove that sin(∞) is the solution set of all possible values for sin(n) where n is any positive number
and since sin(n) can be anything on the interval from -1 <= n <= 1, sin(∞) is all of these values. "all of these values" is [-1,1]
Thanks for reading this goofy proof. I am serious though, I legitimately believe this. This will likely get downvoted for being so unconventional, and since lim(x→∞)[ sin(x) ] does not really approach anything intuitively, I may even be wrong. Tell me what you think, and sorry for all the newline characters that make this answer super long and spaced out.
