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Wed 9 May, 2012 07:55 am
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.
@Termoi,
the limit of 1/(1+e^x) goes as x goes to infinity the function approaches zero--
Rap
@raprap,
As x goes to negative infinity, e^x approaches zero.
1+e^x approaches one.
1/(1+e^x) approaches one as well.
Can 1/(1+e^x) = 1 ? No.
@DrewDad,
Yes, as x goes to negative infininty the function goes to 1. --The function is a horizontal s-curve with asymptotes of y=o to the right and y=1 to the left. Note at x=o, y=1/2.
Rap
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