Question

The value of $\underset{x\to \infty }{\lim }{\left(x+e^x\right)}^{2/x}$ is

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

Answer 1

Answer: (C)

$y=\underset{x\to \infty }{\lim }{\left(x+e^x\right)}^{2/x}({\infty }^0$ form $)$
$\therefore \log y=\underset{x\to \infty }{\lim }2\frac{\log \left(x+e^x\right)}{x}(\frac{\infty }{\infty }.$ form $)$
$=2\cdot \underset{x\to \infty }{\lim }\frac{1+e^x}{x+e^x}$ (Using L'Hospital's Rule)
$=2\cdot \underset{x\to \infty }{\lim }\frac{0+e^x}{1+e^x}$ (Using L'Hospital's Rule)
$=2\cdot \underset{x\to \infty }{\lim }\frac{e^x}{0+e^x}=2$ (Using L'Hospital's Rule)
$\therefore y=e^2$