Does x*sin(x) = θ(x) as x approaches infinity?

Question

I'm trying to determine whether or not xsinx is both O(x) and Ω(x) as x approaches infinity. I know if it fulfills both conditions, it's the same as saying that xsinx = θx, and I'm pretty sure xsinx = O(x), because just looking at them on a graph, the rate of growth of xsinx is clearly not greater than that of x...

picture of y = xsinx and y = x

However, I'm less certain about xsinx = Ω(x) - I know it's a special case because of the oscillating nature but I'm not sure where to go from that.

Can you please define Ω(x) and O(x) ? – Soumyadwip Chanda Sep 15 '20 at 06:04 — Soumyadwip Chanda, Sep 15 '20 at 06:04

score 1 · Answer 1 · answered Sep 15 '20 at 06:05

1

You must use the definition ! Can you establish

$$c_\Omega x\le |x\sin x|\le c_O x$$ for some $c_\Omega,c_O>0$ and $x>x_0$ ?

answered Sep 15 '20 at 06:05

Does x*sin(x) = θ(x) as x approaches infinity?

1 Answers1