# How do you evaluate ((2x-1)/(2x+5))^(2x+3) as x approaches infinity?

Jun 7, 2016

${\lim}_{x \to \infty} {\left(\frac{2 x - 1}{2 x + 5}\right)}^{2 x + 3} = {e}^{- 6}$

#### Explanation:

Making $y = 2 x - 1$ in $f \left(x\right) = {\left(\frac{2 x - 1}{2 x + 5}\right)}^{2 x + 3}$ we get

$g \left(y\right) = {\left(\frac{y}{y + 6}\right)}^{y + 4} = {\left(\frac{y}{y + 6}\right)}^{y} {\left(\frac{y}{y + 6}\right)}^{4}$

but

${\left(\frac{y}{y + 6}\right)}^{y} = {\left(\frac{1}{1 + \frac{6}{y}}\right)}^{y} = {\left({\left(\frac{1}{1 + \frac{6}{y}}\right)}^{\frac{y}{6}}\right)}^{6}$

then

${\lim}_{x \to \infty} f \left(x\right) \equiv {\lim}_{y \to \infty} g \left(y\right)$

and

${\lim}_{y \to \infty} g \left(y\right) = {\left({\lim}_{y \to \infty} {\left(\frac{1}{1 + \frac{6}{y}}\right)}^{\frac{y}{6}}\right)}^{6} {\left({\lim}_{y \to \infty} \left(\frac{1}{1 + \frac{6}{y}}\right)\right)}^{4}$

here
${\lim}_{y \to \infty} \left(\frac{1}{1 + \frac{6}{y}}\right) = 1$

From any Calculus textbook we get

${\lim}_{z \to \infty} {\left(\frac{1}{1 + \frac{1}{z}}\right)}^{z} = {e}^{- 1}$

so

${\lim}_{y \to \infty} g \left(y\right) = {e}^{- 6}$