How do you find the horizontal asymptote for #f(x)=(3e^x)/(2-2e^x) #?

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Answer 1 · 2016-06-27T21:57:57

right sided asymptote is #y = - 3/2 #

left sided asymptote is #y = 0#

for horizontal asymptotes we should look at

#lim_{x \to pm oo} (3e^x)/(2-2e^x)#

#= lim_{x \to pm oo} (3)/(2e^{-x}-2)#

#= lim_{x \to pm oo} (3/2)/(e^{-x}-1)#

which we get by having divided top and bottom by #e^x#

So we focus on what happens to #e^{-x}# as #x \to pm oo#

FOR #x \to + oo#

well, #e^{-x} |_{x \to + oo} = 0#

so # lim_{x \to + oo} (3/2)/(e^{-x}-1) = - 3/2#

but FOR #x \to - oo#

#e^{-x} |_{x \to - oo} #

#= e^{x} |_{x \to + oo} #

#= oo#

so # lim_{x \to - oo} (3/2)/(e^{-x}-1) = 0#