Question

$\underset{x\to \infty }{lim}{\left(\frac{x-2}{x+3}\right)}^{2x}=\underset{x\to \infty }{lim}{(1-\frac{5}{x+3})}^{2x}=e^{-2b}.$ Find b

Answer 1

$\underset{x\to \infty }{lim}{\left(\frac{x-2}{x+3}\right)}^{2x}=\underset{x\to \infty }{lim}{(1-\frac{5}{x+3})}^{2x}$
$=\underset{x\to \infty }{lim}{(1-\frac{5}{x+3})}^{\frac{x+3}{-5}.\frac{-5}{x+3}2x}$
$=e^{\underset{x\to \infty }{lim}(-10).\frac{1}{1+3/x}}$...............$\underset{x\to \infty }{lim}{(1-\frac{1}{x})}^x=e^{-1}$

$=e^{-10}$

$\Rightarrow b=5$