How do you find the vertical, horizontal or slant asymptotes for $\frac{3 e^{x}}{e^{x} - 2}$ $(3e^x) /( e^x-2)$ ?

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Answer 1 · 2016-12-25T06:27:55

Horizontal;) y=0 and y = 3. Vertical: $x = ln 2 = 0.6931$ $x=ln 2=0.6931$ , nearly.

graph{3e^x/(e^x-2) [-20, 20, -10, 10]}

$y = 3 \frac{e^{x}}{e^{x} (1 - 2 e^{- x})} = \frac{3}{1 - 2 e^{- x}}$ $y=3e^x/(e^x(1-2e^(-x)))=3/(1-2e^(-x))$

As $x \to \infty, y \to 3$ $x to oo, y to 3$ . As $x \to - \infty, y \to 0$ $x to -oo, y to 0$ .

So, y - 0 and y = 3 are asymptotes.

By actual division,

$y = 3 + \frac{6}{e^{x} - 2}$ $y = 3+6/(e^x-2)$

Here, as $e^{x} \to 2, y \to \infty$ $e^x to 2, y to oo$ .

So, x = ln 2 gives yet another asymptote.

How do you find the vertical, horizontal or slant asymptotes for (3e^x) /( e^x-2)3exex−2(3e^x) /( e^x-2)?