Why does $\frac{s}{s-1} > \zeta(s) > \frac{1}{s-1}$ imply $\lim_{s \to 1^{+}}(s-1)\zeta(s)=1$?

I am reading the paper Dirichlet's theorem: a real variable approach by Robin Chapman. In this paper, he constructs a proof via real analysis rather than complex analysis that $\zeta(s)$ is convergent if and only if $s>1$. However, this is a standard fact known about $\zeta(s)$. What confuses me is this:

He states as a consequence of the inequality $$\frac{s}{s-1} > \zeta(s) > \frac{1}{s-1},$$ the following limit is true: $$\lim_{s \to 1^{+}}(s-1)\zeta(s)=1.$$ I know this is probably a stupid question, but I'm not that great with limits. I can't quite see where this reasoning is derived from. Is this the case because of the equivalent inequality $$s>(s-1)\zeta(s)>1$$ where $s >1$? If so, how?

Could anyone care to elucidate this rudimentary step in logic for me?

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By the squeeze theorem. –  Najib Idrissi Apr 19 '12 at 11:06
@zulon, So, since $\lim_{s \to 1^{+}}s=1$ and $\lim_{s \to 1^{+}}1=1$, $\lim_{s \to 1^{+}}(s-1)\zeta(s)$ must be $1$? –  000 Apr 19 '12 at 11:13
You're right, that limit is true because of the equivalent inequality you've listed. You just need to apply the squeeze theorem (en.wikipedia.org/wiki/Squeeze_theorem). –  in_wolframAlpha_we_trust Apr 19 '12 at 11:19
@EmileOkada: Oh no, don't delete your answer! I didn't mean that way at all. As long as the OP has it, it doesn't matter who posted the answer. –  Najib Idrissi Apr 19 '12 at 12:08
@EmileOkada, please restore you answer so that I can accept it. :) –  000 Apr 19 '12 at 12:30

This follows from the squeeze theorem. $$\frac{s}{s-1}>\zeta(s)>\frac{1}{s-1}$$ $$\frac{s(s-1)}{s-1}>\zeta(s)(s-1)>\frac{1(s-1)}{s-1}$$ $$s>\zeta(s)(s-1)>1$$ Since $\lim_{s\rightarrow1^+}s=\lim_{s\rightarrow1^+}1=1$ $$\lim_{s\rightarrow1^+}\zeta(s)(s-1)=1$$

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