Question

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

• A. $e$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$0$

Let, $L=\underset{x\to \infty }{lim}{\left(\frac{2x-3}{3x+2}\right)}^x$
$=\underset{x\to \infty }{lim}{\left(\frac{2-3/x}{3+2/x}\right)}^x$
Now, $\underset{x\to \infty }{lim}\left(\frac{2-3/x}{3+2/x}\right)=\frac{2}{3}<1$
Hence $L=\underset{x\to \infty }{lim}{\left(\frac{2-3/x}{3+2/x}\right)}^x=0$