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Analysis is easy

Name: Anonymous 2009-01-03 6:35

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?

Name: Anonymous 2009-01-03 11:37

Thanks.

Another one for anyone who's bored.

Prove \frac{4^{k}}{k!} \rightarrow 0 as n \rightarrow \infty

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Analysis is easy

1 Name: Anonymous 2009-01-03 6:35

3 Name: Anonymous 2009-01-03 11:37

Name: Anonymous 2009-01-03 6:35

Name: Anonymous 2009-01-03 11:37