I have a question concerning the proof of the continuity of $f(x) = \ln x$.
I read in a comment by Pedro Tamaroff to ncmathsadist's answer to this question that this can be proved in two steps:
- first by proving that $f(x)$ is continuous at $1$,
- and then by using $\ln (xy) = \ln(x) + \ln(y)$.
However, I do not really see how to do it.
How does it actually work?
Assume you have shown that $f=\ln$ is continuous at $1$.
We will use the following relation, that holds for any $x_0 > 0$ $$ \ln(x_0+h) = \ln\left(x_0\left(1+\frac{h}{x_0}\right)\right) = \ln x_0 + \ln\left(1+\frac{h}{x_0}\right). \tag{1} $$
For an $(\varepsilon,\delta)$ proof: fix any $x_0>0$, and choose any $\varepsilon > 0$. Let $\delta_1$ be such that $$ \lvert\ln x - \ln 1\rvert \leq \varepsilon $$ whenever $\lvert x-1\rvert \leq \delta_1$ (this exists by continuity of $\ln$ at $1$). Now, set $\delta \stackrel{\rm def}{=} x_0\delta_1$.
For any $x>0$ such that $\lvert x-x_0\rvert \leq \delta$, we have by (1) $$\begin{align} \lvert \ln x - \ln x_0\rvert &= \lvert \ln (x_0+\underbrace{(x-x_0)}_{"h"}) - \ln x_0\rvert = \left\lvert \ln\left( 1+\frac{x-x_0}{x_0}\right)\right\rvert\\ &= \left\lvert \ln\left( 1+\frac{x-x_0}{x_0}\right) - \ln 1\right\rvert \leq \varepsilon \end{align}$$ the last inequality since $\left\lvert\frac{x-x_0}{x_0}\right\rvert \leq \frac{\delta}{x_0} = \frac{x_0\delta_1}{x_0} = \delta_1$.
So $\ln$ is continuous at $x_0$.