Proving that $f(x) = \ln x$ is continuous at $a = 3$.

Question

Proving that $f(x) = ln x$ is continuous at $a = 3$.

The book wrote this solution:

But when I started by $|f(x) - f(3)| < \varepsilon$ to find the appropriate $\delta$, I arrived at $ 3e^{-\varepsilon} -3 < x-3 < 3e^\varepsilon -3$, I do not understand why he take delta the minimum of the left hand side and right hand side of $x-3$, why he did not take directly that $\delta = 3e^\varepsilon -3$? could anyone explain this for me please?

Answer 1

$\big|\ln x-\ln 3\big|=\big|\ln\left(\dfrac{x}{3}\right)\big|\lt\varepsilon$ whenever $-\varepsilon\lt\ln\left(\dfrac{x}{3}\right)\lt\varepsilon\implies e^{-\varepsilon}\lt\dfrac{x}{3}\lt e^{\varepsilon}\implies 3e^{-\varepsilon}\lt x\lt 3e^{\varepsilon}.$

Thus $\big|\ln x-\ln 3\big|\lt\varepsilon$ whenever $|x-3|\lt\delta$

where $\delta$=$\min(3-3e^{-\varepsilon},3e^{\varepsilon}-3).$

$\varepsilon\gt 0\implies e^{\varepsilon}\gt 1\implies \dfrac{1}{e^{\varepsilon}}\lt 1\implies \dfrac{e^{\varepsilon}-1}{e^{\varepsilon}}\lt e^{\varepsilon}-1\implies 1-e^{-\varepsilon}\lt e^{\varepsilon}-1\implies 3-3e^{-\varepsilon}\lt 3e^{\varepsilon}-3\implies \min(3-3e^{-\varepsilon},3e^{\varepsilon}-3)=3-3e^{-\varepsilon}.$

$\therefore \big|\ln x-\ln 3\big|\lt\varepsilon$ whenever $|x-3|\lt3e^{-\varepsilon}-3$

Hence $\ln x$ is continuous at $x=3$.