Question

Let $f\left(x\right)=\{\frac{\log (1+x{)}^{1+x}}{x^2}-\frac{1}{x}\}$. Then the value of $f\left(0\right)$ so that the function $f$ is continuous is

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{2}$

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

Using L-hospital's rule,
$=\underset{x\to 0}{lim}\frac{log(1+x)+1-1}{2x}$
$=\underset{x\to 0}{lim}\frac{log(1+x)}{x}=\frac{1}{2}$