Question

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

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

Answer 1

Answer: (C)

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