Question

For $x \in R, \displaystyle\lim _{x \rightarrow \infty}\left(\frac{x-3}{x+2}\right)^{x}$ is equal to :

• A. $e$
• B. $e^{-1}$
• C. $e^{-5}$
• D. $e^5$

Answer 1

Answer: (C)

$\displaystyle\lim _{x \rightarrow \infty}\left(\frac{x-3}{x+2}\right)^{x}$
$=\displaystyle\lim _{x \rightarrow \infty}\left[1-\frac{5}{x+2}\right]^{x}$
$=\displaystyle\lim _{x \rightarrow \infty}\left[\left(1 +\left(\frac{-5}{x+2}\right)\right)^{1/\left(\frac{-5}{x+2}\right)}\right]^{\left(\frac{-5x}{x+2}\right)}$
$=e^{x \rightarrow \infty}\left(-\frac{5}{1+2 / x}\right)$
$=e^{-5}$
Alternative Solution :
$ \displaystyle\lim _{x \rightarrow \infty}\left(\frac{x-3}{x+2}\right)^{x} $
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\left(1-\frac{3}{x}\right)^{x}}{\left(1+\frac{2}{x}\right)^{x}}$
$= \frac{e^{-3}}{e^{2}}=e^{-5}$