Question

The value of $\underset{x\to 0^{-}}{lim}\frac{\ln (1+\{x\})}{\left\{x\right\}}$ is (where {$x$} denotes the fractional part of x)

• A. $1$
• B. $ln1$
• C. $ln2$
• D. $2$

Answer 1

The correct option is C $ln2$

$\underset{x\to 0^{-}}{lim}\frac{ln(1+\{x\})}{\left\{x\right\}}$
$\underset{h\to 0}{lim}\frac{\ln (1+\{(-h)\left\}\right)}{\{-h\}}$
$\underset{h\to 0}{lim}\frac{\ln (1+(1-h))}{1-h}$
$=\ln 2$