Question

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

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

Answer 1

Answer: (C)

$-\frac{e}{2}$

${lim}_{x\to 0}\frac{(1+x{)}^{1/x}-e}{x}={lim}_{x\to 0}\frac{e^{(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+..)}-e}{x}$
Applying L'Hopital rule
$={lim}_{x\to 0}\frac{{(-\frac{1}{2}+\frac{2x}{3}-\frac{{3x}^2}{4}+..)e}^{(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+..)}}{1}=-\frac{e}{2}$
Hence, option 'C' is correct.