We set

$$a_{m}=\max_{x\in X^{\mathrm{an}}}\left\{\log\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{m}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{m}}(x)}\right\}.$$

Then $a_{m+m^{\prime}}\leq a_{m}+a_{m^{\prime}}$ by (3) in Lemma 3.5, and hence $\lim_{m\to\infty}a_{m}/m=\inf\{a_{m}/m\}$ by Fekete’s lemma. For $\epsilon>0$, there is $e_{n}$ such that

$$e^{-e_{n}\epsilon}|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)\leq|\raisebox{1.72218pt}{.}|_{h_{n}}(x)\leq e^{e_{n}\epsilon}|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)$$

for all $x\in X^{\mathrm{an}}$, where $h_{n}=\big\{|\raisebox{1.72218pt}{.}|^{\operatorname{quot}}_{(V_{n},\|\raisebox{1.20552pt}{.}\|_{n})}(x)\big\}_{x\in X^{\operatorname{an}}}$. Thus

$$e^{-e_{n}\epsilon}\|\raisebox{1.72218pt}{.}\|_{h^{e_{n}}}\leq\|\raisebox{1.72218pt}{.}\|_{h_{n}}\leq e^{e_{n}\epsilon}\|\raisebox{1.72218pt}{.}\|_{h^{e_{n}}},$$

so that $e^{-e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)\leq|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h_{n}}(x)\leq e^{e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)$. Thus, by Proposition 3.6,

$$e^{-e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)\leq|\raisebox{1.72218pt}{.}|_{h_{n}}(x)\leq e^{e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x).$$

Therefore,

$$1\leq\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)}=\frac{|\raisebox{1.72218pt}{.}|_{h_{n}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)}\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)}{|\raisebox{1.72218pt}{.}|_{h_{n}}(x)}\leq e^{2e_{n}\epsilon},$$

that is, $0\leq a_{e_{n}}/e_{n}\leq 2\epsilon$, and hence $0\leq\lim_{m\to\infty}a_{m}/m\leq 2\epsilon$, as required. ∎