Question

$\underset{h\to 0}{lim}\frac{-\log (1+2h)+2\log (1+h)}{h^2}$ is equal to

Answer 1

The required limit is equal to
$\underset{h\to 0}{lim}\frac{1}{h^2}[-\log (1+2h)+2log(1+h\left)\right]$

$=-\underset{h\to 0}{lim}\frac{1}{h^2}\log \frac{1+2h}{(1+h{)}^2}$

$=-\underset{h\to 0}{lim}\frac{\log (1-\frac{h^2}{(1+h{)}^2})}{-\frac{h^2}{(1+h{)}^2}}\times -\frac{1}{(1+h{)}^2}=1$