Answer
$\kappa (x)$ has maximum at $(\sqrt{\frac{1}{2}},\frac{1}{2} \ln \frac{1}{2})$
and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$
Work Step by Step
Given: $y=\ln x$
Consider $f(x)=y=\ln x$
In order to find the curvature we will have to use formula 11, such that
$\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$
$y'=\frac{1}{x}$
and $y''=-\frac{1}{x^2}$
$\kappa(x)=\dfrac{|-\frac{1}{x^2})|}{[1+(\frac{1}{x})^2]^{3/2}}=\dfrac{\frac{1}{x^2}}{[1+\frac{1}{x^2}]^{3/2}}=\dfrac{x}{(1+x^2)^{3/2}}$
$\kappa'(x)=\dfrac{1-2x^2}{(1+x^2)^{3/2}}$
$\kappa (x)$ has maximum at $(\sqrt{\frac{1}{2}},\frac{1}{2} \ln \frac{1}{2})$
and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$