Question

$\underset{x\to 0}{lim}{\left(\frac{1+x}{1-x}\right)}^{1/x}$ equals

• A. $e$
• B. $e^2$
• C. $\frac{1}{e}$
• D. $\frac{1}{e^2}$

Answer 1

Answer: (B)

$e^2$

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