Question

The value of $\underset{x\to \infty }{lim}{\left[\frac{e}{{(1+\frac{1}{x})}^x}\right]}^x$ is given by?

• A. $e$
• B. $e^{-1}$
• C. $e^{1/2}$
• D. $e^{-1/2}$

Answer 1

The correct option is C $e^{1/2}$

$\underset{x\to \infty }{lim}{\left[\frac{e}{{(1+\frac{1}{x})}^x}\right]}^x=\underset{y\to 0}{lim}{\left[\frac{e}{{(1+y)}^{\frac{1}{y}}}\right]}^{\frac{1}{y}}$$=\underset{y\to 0}{lim}{e}^{\frac{1}{y}\ln \left[\frac{e}{{(1+y)}^{\frac{1}{y}}}\right]}$

Now, we have,

$\underset{y\to 0}{lim}\frac{1}{y}\ln \left[\frac{e}{{(1+y)}^{\frac{1}{y}}}\right]=\underset{y\to 0}{lim}\frac{\ln e-\frac{1}{y}\ln (1+y)}{y}$$=\underset{y\to 0}{lim}\frac{y-\ln (1+y)}{y^2}$

$=\underset{y\to 0}{lim}\frac{y-(y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+...)}{y^2}$

$=\underset{y\to 0}{lim}\frac{1}{2}-\frac{y}{3}+\frac{y^2}{4}-...=\frac{1}{2}$

Hence, the required limit is $e^{1/2}.$