I think you're approaching the question in the wrong way. The whole point is that you can show that for $\Re(s) > 1$, $$\Phi(s) = \sum_{p}{\frac{\log p}{p^s}} = \frac{1}{s - 1} + E(s),$$ where the function $E(s)$ is holomorphic meromorphic on the larger open half-plane $\Re(s) > 1/2$ ; with poles possibly at the zeroes of $\zeta(s)$; in fact, Zagier shows that $$E(s) = - \frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s - 1} - \sum_{p}{\frac{\log p}{p^s (p^s - 1)}}.$$
Basically, this is saying that $\Phi(s)$ is meromorphic on an open neighbourhood of $\Re(s) > 1/2$ \geq 1$with a simple pole at$s = 1$, and the expansion above shows that the residue of$\Phi(s)$at$s = 1$is equal to$1$. This is precisely the same as saying that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1.$$ Indeed, we have that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = \lim_{\varepsilon \to 0} \frac{\varepsilon}{1 + \varepsilon - 1} + \lim_{\varepsilon \to 0} \varepsilon E(1 + \varepsilon),$$ and the first limit tends to$1$(obviously) while the second limit tends to zero (as$E(1 + \varepsilon)$tends to something finite). If you don't understand this method at all (i.e. all about meromorphic extensions of functions, poles, and residues), then this is probably due to a lack of background in complex analysis. Seeing as this proof of the prime number theorem is all about complex analysis, I'd recommend reading up on all these basics beforehand. 1 I think you're approaching the question in the wrong way. The whole point is that you can show that for$\Re(s) > 1$, $$\Phi(s) = \sum_{p}{\frac{\log p}{p^s}} = \frac{1}{s - 1} + E(s),$$ where the function$E(s)$is holomorphic on the larger half-plane$\Re(s) > 1/2$; in fact, Zagier shows that $$E(s) = - \frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s - 1} - \sum_{p}{\frac{\log p}{p^s (p^s - 1)}}.$$ Basically, this is saying that$\Phi(s)$is meromorphic on$\Re(s) > 1/2$with a simple pole at$s = 1$, and the expansion above shows that the residue of$\Phi(s)$at$s = 1$is equal to$1$. This is precisely the same as saying that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1.$$ Indeed, we have that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = \lim_{\varepsilon \to 0} \frac{\varepsilon}{1 + \varepsilon - 1} + \lim_{\varepsilon \to 0} \varepsilon E(1 + \varepsilon),$$ and the first limit tends to$1$(obviously) while the second limit tends to zero (as$E(1 + \varepsilon)\$ tends to something finite).