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

1 Answer
Jan 21, 2017

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.