Answer
The horizontal asymptotes are:
$y = 0~~$ (as $x \to -\infty$)
$y = 2~~$ (as $x \to \infty$)
$x=ln(5)~~$ is a vertical asymptote.
Work Step by Step
Horizontal asymptotes:
$\lim\limits_{x \to -\infty}\frac{2e^x}{e^x-5} = \lim\limits_{x \to \infty}\frac{2/e^x}{1/e^x-5} = \frac{0}{0-5} = 0$
$\lim\limits_{x \to \infty}\frac{2e^x}{e^x-5} = 2$
The horizontal asymptotes are $y = 0~~$ (as $x \to -\infty$) and $y = 2~~$ (as $x \to \infty$)
Vertical asymptotes.
$y = \frac{2e^x}{e^x-5}$
We can find $x$ when $e^x-5 = 0$:
$e^x-5 = 0$
$e^x = 5$
$x = \ln(5)$
Thus $~x=ln(5)~~$ is a vertical asymptote.
