Question

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

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

Answer 1

The correct option is C $e^{-5}$

${lim}_{x\to \infty }{\left(\frac{x-3}{x+2}\right)}^x={lim}_{x\to \infty }{(1-\frac{5}{x+2})}^x={lim}_{x\to \infty }{(1-\frac{5}{x+2})}^{-\frac{x+2}{5}\times \frac{-5}{x+2}\times x}={lim}_{x\to \infty }{(1-\frac{5}{x+2})}^{-\frac{x+2}{5}\times (-5)\times \frac{1}{1+2/x}\times x}=e^{-5\times 1}=e^{-5}$
Hence, option 'C' is correct.