Answer
$\lim\limits_{x \to \infty}(\frac{2x-3}{2x+5})^{2x+1} = \frac{1}{e^{8}}$
Work Step by Step
Let $~~y = \lim\limits_{x \to \infty}(\frac{2x-3}{2x+5})^{2x+1}$
$ln~y = \lim\limits_{x \to \infty}ln (\frac{2x-3}{2x+5})^{2x+1}$
$ln~y = \lim\limits_{x \to \infty}(2x+1)~ln (\frac{2x-3}{2x+5})$
$ln~y = \lim\limits_{x \to \infty}\frac{ln (\frac{2x-3}{2x+5}~)}{(2x+1)^{-1}} = \frac{0}{0}$
We can use L'Hospital's Rule:
$ln~y = \lim\limits_{x \to \infty}\frac{\frac{2x+5}{2x-3}~\cdot ~\frac{2(2x+5)-2(2x-3)}{(2x+5)^2}}{-2(2x+1)^{-2}}$
$ln~y = \lim\limits_{x \to \infty}\frac{\frac{16}{(2x-3)(2x+5)}}{-2(2x+1)^{-2}}$
$ln~y = \lim\limits_{x \to \infty}\frac{-8(2x+1)^2}{(2x-3)(2x+5)}$
$ln~y = \lim\limits_{x \to \infty}\frac{-8(4x^2+4x+1)}{4x^2+4x-15}$
$ln~y = \lim\limits_{x \to \infty}\frac{-32x^2-32x-8}{4x^2+4x-15}$
$ln~y = \lim\limits_{x \to \infty}\frac{-32-32/x-8/x^2}{4+4/x-15/x^2}$
$ln~y = -\frac{32}{4}$
$ln~y = -8$
$y = e^{-8}$
$y = \frac{1}{e^{8}}$
