Math
posted by Lawn .
Hi everyone,
Sorry to bother you, I was wondering, why is it that for the graph of 1/(1+e^x), 1 is the horizontal asymptote?
I had thought the HA should be something that makes the reciprocal function undefined, so I kind of thought the HA should be 1.
Please explain simply, thank you very much! Really appreciate the help.

were you allowed to use a graphing calculator? because if you were, you would be able to find the HA by looking at the graph

The horizontal asymptote indicates what happens to y when x gets large.
As x gets huge negative, e^x > 0, so we wind up with y = 1/1 = 1
As x gets huge positive, e^x > oo, so we wind up with y=1/oo = 0
So, there are two horizontal asymptotes, at y=0 and y=1
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