Question

Evaluate : ${lim}_{x\to 0}\frac{(1+x{)}^{1/x}-e}{x}=?$

• A. $1$
• B. $e/2$
• C. $-e/2$
• D. $2/e$

Answer 1

The correct option is C $-e/2$

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

$=\underset{x\to 0}{lt}\frac{{(1+x)}^{1/x}[\frac{x}{1+x}-ln(x+1)]/x^2}{1}$

$=\underset{x\to 0}{lt}\frac{e(\frac{(1+x)-x}{{(1+x)}^2}-\frac{1}{(x+1)})}{2x}$

$=e\underset{x\to 0}{lt}\frac{1-(x+1)}{{(x+1)}^2\times 2x}$

$e\underset{x\to 0}{lt}\frac{x}{{(x+1)}^{2}2x}$

$=-e\underset{x\to 0}{lt}\frac{x}{{(x+1)}^2\times 2x}$

$=\frac{-e}{2}$