Analysis is easy

Let S = { \frac{x-2}{x+3} \mid x \in \Re, x \geq 0 }

Prove that Sup(S) = 1 and Inf(S) = -\frac{2}{3}

It's easy to see it's true but how to prove it fairly rigorously?

>>9
Don't bother read that.  Real analysis is just like calculus, only more boring.  Complex analysis is where it starts to get (slightly) interesting.

http://en.wikipedia.org/wiki/Complex_analysis