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?

hey, im a newfag in high school calculus. Excuse my ignorance, but what kind of math are you chaps doing here? I'm intrigued