Question

How do you evaluate the limit `sin(x^2)` as x approaches `oo`?

Answer 1

This limit doesn't exist. Explanation below.

Explanation:

`sin x` constantly changes, taking values between `1` and `-1` and back ("oscillates") as `x` increases. `sin x^2` also has this oscillation effect, only the graph is a little different, and in fact, it oscillates more rapidly as `x` approaches infinity, since the rate of change of `x^2` also increases. Because of that, since `sin x^2` neither goes unbounded, nor does it approach a set value as `x` approaches infinity, its limit at infinity doesn't exist.