Question

What is the limit of `Abs(sinx)` as x approaches infinity?

Answer 1

There is no limit.

Explanation:

Consider what happens as `x` increases without bound.

For any value of `x`, no matter how great, there are values of the form `pin` and `pi/2+pin` to the right of `x`.
Therefore there are places to the right where `abs(sinx) = 0` and places to the right where `abs(sinx) = 1`.

The values of `abs(sinx)` cannot be getting closer and closer to a single value.

A more rigorous argument can be based on the following.

If someone claims that the limit is `L`, they must show that for every positive `epsilon`, there is a number `M` such that for all `x > M`, we have `abs(abs(sinx)-L) < epsilon`

I claim that for `epsilon <= 1/2` this is not possible.