Question

The value of $\underset{x\to 0}{lim}\frac{(1+x{)}^{1/x}-e}{x}$ is

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

Answer 1

The correct option is A $-\frac{e}{2}$

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

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

$=\underset{x\to 0}{lim}\frac{e^{1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}...}-e}{x}(\because \log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3})$

$=e\underset{x\to 0}{lim}\frac{e^{-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}...}-1}{x}$

$=e\underset{x\to 0}{lim}\frac{e^{-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}...}-1}{-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}...}\times \frac{-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}...}{x}$

$=e\left(1\right)(-\frac{1}{2})$
$=-\frac{e}{2}$

Alternate Solution
$\underset{x\to 0}{lim}\frac{(1+x{)}^{1/x}-e}{x}=\underset{x\to 0}{lim}\frac{e^{\frac{1}{x}\ln (1+x)}-e}{x}\left[\frac{0}{0}\text{form}\right]\text{Applying L'Hopital's Rule}=\underset{x\to 0}{lim}\frac{e^{\frac{1}{x}\ln (1+x)}[\frac{1}{x}\times \frac{1}{1+x}-\frac{\ln (1+x)}{x^2}]}{1}=\underset{x\to 0}{lim}(1+x{)}^{1/x}\times \underset{x\to 0}{lim}\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}=e\times \underset{x\to 0}{lim}\frac{\frac{1}{1+x}-\frac{x}{(1+x{)}^2}-\frac{1}{(1+x)}}{2x}=e\times \underset{x\to 0}{lim}\frac{-1}{2(1+x{)}^2}=-\frac{e}{2}$