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?

>>3
This stuff is pretty simple, what are you having trouble with?
Is it merely how to show the things "rigorously"?

In this case there's quite a few ways to do it.

I'd probably just say

4^k/k! < 4/k*(4^k-1/(k-1)!) < 4/8*(4^k-1/(k-1)!) for k>8

So if you let A_k = 4^k/k! you can see that A_k < 1/2*A_(k-1)


so as k-> infinity A_k < 1/2^n*A_8 -> 0 so done.

Obviously that could be tidied up, wanted to walk you through it