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July 30, 2014

July 30, 2014

Posted by **Lawn** on Wednesday, May 9, 2012 at 9:49am.

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.

- Math -
**Anon**, Wednesday, May 9, 2012 at 10:08amwere you allowed to use a graphing calculator? because if you were, you would be able to find the HA by looking at the graph

- Math -
**Steve**, Wednesday, May 9, 2012 at 1:15pmThe 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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