Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

The correct option is B $-1$

Substituting gives us indeterminate form.
So, applying L'Hospitals Rule, $L=\underset{x\to 0}{lim}\frac{\frac{2}{1+2h}-\frac{2}{1+h}}{2h}$
Substituting gives us indeterminate form so, applying L'Hospitals Rule again,
$=\underset{x\to 0}{lim}\frac{\frac{-4}{(1+2h{)}^2}+\frac{2}{(1+h{)}^2}}{2}$
$=\frac{-4+2}{2}$ (direct substitution)
$=-1$