How do you find the vertical, horizontal or slant asymptotes for #(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#, nearly.

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

#y=3e^x/(e^x(1-2e^(-x)))=3/(1-2e^(-x))#

As #x to oo, y to 3#. As #x to -oo, y to 0#.

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

By actual division,

#y = 3+6/(e^x-2)#

Here, as #e^x to 2, y to oo#.

So, x = ln 2 gives yet another asymptote.