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? 

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