Question

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

• A. $\frac{1}{1-e}$
• B. $0$
• C. $\frac{1}{e^{1-e}}$
• D. Does not exist

Answer 1

Answer: (D)

Does not exist

$\underset{x\to \infty }{lim}{(\frac{1}{e}-\frac{x}{1+x})}^x=\underset{x\to \infty }{lim}{(\frac{1}{e}-\frac{1}{\frac{1}{x}+1})}^x={(\frac{1}{e}-1)}^{\infty }$$={\text{( some negative value )}}^{\infty }$which is not defined as base is negative