What are all horizontal asymptotes of the graph #y=(5+2^x)/(1-2^x)# ?

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Answer 1 · 2014-09-22T04:09:35

Let us find limits at infinity.

#lim_{x to +infty}{5+2^x}/{1-2^x}#

by dividing the numerator and the denominator by #2^x#,

#=lim_{x to +infty}{5/2^x+1}/{1/2^x-1}={0+1}/{0-1}=-1#

and

#lim_{x to -infty}{5+2^x}/{1-2^x}={5+0}/{1-0}=5#

Hence, its horizontal asymptotes are

#y=-1# and #y=5#

They look like this:
enter image source here