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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.2012-04-19
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    @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 $1$?2012-04-19
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    You're right, that limit is true because of the equivalent inequality you've listed. You just need to apply the squeeze theorem (http://en.wikipedia.org/wiki/Squeeze_theorem).2012-04-19
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    @zulon: maybe you can post an answer to settle this?2012-04-19
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    @J.M.: Looks like I'm too late :)2012-04-19
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    @zulon, I was awaiting you to post an answer. I figured it would be more respectful to allow you to be the accepted answer, since you were the first to remark on this question and Emile Okada was not. Would you prefer posting an answer for me to accept or me accepting Emile's answer?2012-04-19
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    @zulon sorry about that. I didn't notice the comments until now. I will be happy to delete my answer if you post your own.2012-04-19
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    @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.2012-04-19
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    @EmileOkada, please restore you answer so that I can accept it. :)2012-04-19

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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 $\lim_{s\rightarrow1^+}s=\lim_{s\rightarrow1^+}1=1$ $$\lim_{s\rightarrow1^+}\zeta(s)(s-1)=1$$