Why would $\forall x\log(x) = 0 \implies 2^\frac{1}{n} - 1 \leq \frac{\epsilon}{n}$ for large $n$? I'm reading a calculus text which used this in a reductio to prove the log function is nontrivial and I don't understand the implication. Note: the text defines $\log(b)$ as the derivative of $\exp_b(x)$ at 0, so no other definitions can be used in the answers.
EDIT: Heres is the screen grab from the book pdf:
I use the definition that the derivative of a differentiable function $f$ at $x \in \mathbb{R} $ is:
$$ f'(x) = \lim_{h \rightarrow 0 } \frac{f(x+h) - f(x)}{h} $$
Specifically this implies:
$$ \forall \epsilon > 0 \ \ \ \ \ \exists n \in \mathbb{N} \ \ \ \left| \frac{f(x+1/n) - f(x)}{(1/n)} - f'(x) \right| < \epsilon $$
Define $ \log(b) $ as $ \frac{d}{dx}(b^x) $ evaluated at $x = 0 $, as in your question.
If $\forall y \in \mathbb{R} \ \ \log(y) = 0 $, then certainly $\log(2) = 0$ .
Then by the limit definition of the derivative:
$$ \forall \epsilon > 0 \ \exists n \in \mathbb{N} \ \ \ \ \ \ n(2^{1/n} - 1)= \frac{2^{0+1/n} - 2^0}{(1/n)} - \log(2) \le \left| \frac{2^{0+1/n} - \ 2^0}{(1/n)} -\log(2) \right| \le \epsilon $$
and then divide by $n$