Question

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

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

Answer 1

The correct option is C $e^{\frac{e}{1-e}}$

$\underset{x\to \infty }{lim}{\left[\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right]}^x=e^{\underset{x\to \infty }{lim}[\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)-1]x}$
$=e^{\underset{x\to \infty }{lim}\left[\frac{1}{1-e}-\frac{e}{1-e}\frac{x}{1+x}-1\right]x}$
$=e^{\underset{x\to \infty }{lim}\left[\frac{e}{1-e}-\frac{e}{1-e}\frac{x}{1+x}\right]x}$
$=e^{\underset{x\to \infty }{lim}\frac{e}{1-e}[1-\frac{x}{1+x}]x}$
$=e^{\frac{e}{1-e}\underset{x\to \infty }{lim}\frac{x}{1+x}}$
$=e^{\frac{e}{1-e}\underset{x\to \infty }{lim}\frac{1}{1+\frac{1}{x}}}$
$=e^{\frac{e}{1-e}}$
Hence, option 'C' is correct.