I'm just trying to do a basic $\delta - \epsilon$ proof, but I'm getting lost finding delta. The function is $f(x) = -\log_2(\frac{1-x}{x})$ so I get that I need $|\log_2 \frac{x-xv}{v-xv}| \lt \epsilon$ so $\frac{x-xv}{v-xv} \lt 2^\epsilon$, but I'm struggling with getting the variables out of the way and manipulating this to have $\delta(v)$
EDIT: $x,v \in (0,1), f: (0,1) \rightarrow \mathbb{R}$
Don't forget you really need the absolute value bounded, so you have to consider the lower bound too:
$$ \begin{aligned} 2^{-\epsilon}<\frac{x(1-v)}{v(1-x)}<2^{\epsilon}&\iff 2^{-\epsilon}\frac{1-x}{x}<\frac{1}{v}-1<2^{\epsilon}\frac{1-x}{x}\\[10pt] &\iff 2^{-\epsilon}\frac{1-x}{x}+1<\frac{1}{v}<2^{\epsilon}\frac{1-x}{x}+1\\[10pt] &\iff\frac{2^{-\epsilon}-2^{-\epsilon}x+x}{x}<\frac{1}{v}<\frac{2^{\epsilon}-2^{\epsilon}x+x}{x}\\[10pt] &\iff\frac{x}{2^{\epsilon}-2^{\epsilon}x+x}<v<\frac{x}{2^{-\epsilon}-2^{-\epsilon}x+x}. \end{aligned} $$