Question

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

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

Answer 1

Answer: (C)

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