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?

For any e > 0 you exhibit an x in R such that x-2/x+3 > 1-e

and similarly a y in R such that y-2/y+3 < -2/3 + e

That is if you know that 1 is an upper bound and -2/3 is a lower bound already, which you should also show, but that's basic.

so for example:

let x = n a natural number

then n-2/n+3 > n-2/n = 1 - 2/n  so that's Sup(S) done, as obviously x-2 < x+3 => x-2/x+3 < 1

Do the rest yourself