Sine of Infinity?

"$\sin \infty$" (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.

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.

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.