Question

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

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

Answer 1

The correct option is B $\frac{e}{2}$

$\underset{x\to 0}{lim}\frac{e-{(1+x)}^{\frac{1}{x}}}{x}=\underset{x\to 0}{lim}\frac{e-e^{\frac{\ln (1+x)}{x}}-1}{x}$
$=\underset{x\to 0}{lim}e.\frac{e^{\frac{\ln (1+x)-x}{x}}-1}{x}=\underset{x\to 0}{lim}-e\frac{e^y-1}{y}.\frac{\ln (1+x)-x}{x^2}$
Putting $\frac{\ln (1+x)-x}{x}=y$
Now we have $\underset{x\to 0}{lim}\frac{\ln (1+x)-x}{x}=\underset{x\to 0}{lim}\frac{\frac{1}{1+x}-1}{1}=0$
Also $\underset{x\to 0}{lim}\frac{\ln (1+x)-x}{x^2}=\underset{x\to 0}{lim}\frac{\frac{1}{1+x}-1}{2x}=\underset{x\to 0}{lim}\frac{-1}{2(1+x)}=\frac{-1}{2}$
Hence the required limit is
$=-e\underset{y\to 0}{lim}\frac{e^y-1}{y}\underset{x\to 0}{lim}\frac{\ln (1+x)-x}{x^2}=-e.1.\frac{-1}{2}=\frac{e}{2}$