Epsilon delta proof check I tried to prove:
\begin{align*}
&\lim_{x \to c} \log_b(x)=\log_b(c)\\
&\log_b(c) - \epsilon < \log_b(x) < \log_b(c) + \epsilon \\
&b^{\log_b(c) - \epsilon }<b^{\log_b(x)}<b^{\log_b(c) + \epsilon} \\
&c b^{- \epsilon }<x<c b^{\epsilon} \\
&c b^{- \epsilon }-c<x-c<c b^{\epsilon}-c \\
&\delta \leq \min\{c b^{- \epsilon }-c,c b^{\epsilon}-c\}=  c\min\{b^{- \epsilon }-1,b^{\epsilon}-1\}\\ 
\end{align*}
$$c\min\{b^{- \epsilon }-1,b^{\epsilon}-1\}=c(b^{-\epsilon}-1)$$
and as said in $\min$ in epsilon-delta the minimum should be used
However:
\begin{align*}
b^{-\epsilon} -1&>0 \\
b^{-\epsilon} &> 1 \\
\log(b^{-\epsilon}) &>\log(1) \\
-\epsilon \log(b) &>0\\
-\epsilon&>0\\
\epsilon &< 0
\end{align*}
(contradiction)
So do I use:
$$\delta \leq b^\epsilon -1$$
 A: You have missed a sign change. But nice work (other than that one error) and good for you for checking it the way you did.
Notice that when you actually do the proof, you will have a step where you assert that
$$ \lvert x - c \rvert < \delta. $$
That means
$$ -\delta < x - c < \delta. $$
To emphasize this: the thing on the far left is not $\delta$;
it is $-\delta.$
So your two candidates for $\delta are actually $c b^{\epsilon} - c$ (from the far right) and $-(c b^{-\epsilon} - c) = c - b^{-\epsilon}$ (negative of the thing on the far left).
Therefore you must have
\begin{align}
\delta &\leq c(b^{\epsilon} - 1) && \text{(otherwise you let in bad values of $x$ on the right),}\\
\delta &\leq c(1 - b^{-\epsilon}) && \text{(otherwise you let in bad values of $x$ on the left).}\\
\end{align}
Note that if you suppose $1 - b^{-\epsilon} > 0$ then it implies $\epsilon > 0.$
Now you just need to decide which is smaller,
$b^{\epsilon} - 1$ or $1 - b^{-\epsilon}.$
(Hint: $1 - b^{-\epsilon} = b^{-\epsilon}(b^{\epsilon} - 1).$)
