The value of $\underset{h\to 0}{lim} \frac{\ln (1+2h)-2\ln (1+h)}{h^2}$, is

Solution:

We have, 

$\begin{array}{l}\underset{h\to 0}{lim} \frac{\log (1+2h)-2\log (1+h)}{h^2}\\ =\underset{h\to 0}{lim} \frac{\log \left\{\frac{1+2h}{(1+h{)}^2}\right\}}{h^2}=-\underset{h\to 0}{lim} \frac{\log \left\{\frac{(1+h{)}^2}{1+2h}\right\}}{h^2}\\ =-\underset{h\to 0}{lim} \frac{\log \left\{1+\frac{h^2}{1+2h}\right\}}{\left\{\frac{h^2}{1+2h}\right\}(1+2h)}=-1=-1.\end{array}$