Question

$\underset{x\to \infty }{lim}{\left[\left(\frac{e}{1-e}\right)(\frac{1}{e}-\frac{x}{1+x})\right]}^x=$

• A. $e^{(1-e)}$
• B. $e^{\left(\frac{1-e}{e}\right)}$
• C. $e^{\left(\frac{e}{1-e}\right)}$
• D. $e^{\frac{1+e}{e}}$

Answer 1

The correct option is C $e^{\left(\frac{e}{1-e}\right)}$

$\underset{x\to \infty }{lim}{\left[\left(\frac{e}{1-e}\right)(\frac{1}{e}-\frac{x}{1+x})\right]}^x,\left[1^{\infty }\text{Form}\right]$
$=e^{\underset{x\to \infty }{lim}[\left(\frac{e}{1-e}\right)(\frac{1}{e}-\frac{x}{1+x})-1]x}$
$=e^{\underset{x\to \infty }{lim}[\frac{1}{1-e}-\frac{e}{1-e}\frac{x}{1+x}-1]x}$
$=e^{\underset{x\to \infty }{lim}[\frac{e}{1-e}-\frac{e}{1-e}\frac{x}{1+x}]x}$
$=e^{\underset{x\to \infty }{lim}\frac{e}{1-e}[1-\frac{x}{1+x}]x}$
$=e^{\underset{x\to \infty }{lim}\frac{e}{1-e}\left[\frac{x}{1+x}\right]}$
$=e^{\frac{e}{1-e}}$