Personal Scratchpad

[bo198214]

Ah, perhaps I should use another epsilon instead of zero there, with the limit as h goes to epsilon, and taking a limit as epsilon goes to 0 but is still formally NOT zero. It doesn't affect the final result. Seems a bit pedantic as well, but apparently that's part of doing proofs.

Pedantic or not, I have to understand what you write and in this case it is not that clear that I would approve it.
The proper way of doing the derivation probably may probably be:
There is a c>0 such that for each \( \epsilon> \) there is a \( h>0 \) such that
\( c-\eps < \frac{f(x+h)-f(x)}{h} < c+\eps \) or equivalently
\( \left|\frac{f(x+h)-f(x)}{h} - c \right| < \eps \).
Then you can do the rearrangements without limits. For example
\( (c-\eps)h+f(x) < f(x+h) < (c+\eps)h+f(x) \).

And you can perhaps use the continuity and strict monotony of the log function, namely for each \( \eps_2>0 \) there is a \( \delta>0 \) such that
\( \log(x) < \log(x+\delta) < \log(x) + \eps_2 \)
where you let \( \eps_2=(c+\eps)h \) or so. But in the moment I am to lazy to apply it myself.