Question

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

• A. $e$
• B. $e^{-1}$
• C. $e^{-2}$
• D. none of these

Answer 1

Answer: (C)

$e^{-2}$

${lim}_{x\to \infty }{\left(\frac{x-1}{x+1}\right)}^x={lim}_{x\to \infty }{(1-\frac{2}{x+1})}^x={lim}_{x\to \infty }{(1-\frac{2}{x+1})}^{-\frac{x+1}{2}\times \frac{-2}{x+1}\times x}={lim}_{x\to \infty }{(1-\frac{2}{x+2})}^{-\frac{x+1}{2}\times -2\times \frac{1}{1+1/x}}=e^{-2\times 1}=e^{-2}$
Hence, option 'C' is correct.