find the limit x--> -infinity (e^x-e^(-x))/(e^x+e^(-x))
2007-10-18 17:09:48 · 4 answers · asked by kris 1 in Science & Mathematics ➔ Mathematics
limit x--> -∞ (e^x-e^(-x))/(e^x+e^(-x))
Dividing numerator and denominator by e^x
limit
= limit x--> -∞ (1 - e^(-2x)) / (1 + e^(-2x))
= limit x--> -∞ [ {1 - 1 / e^(2x)} / {1 + 1 / e^(2x)} ]
= 1 (because limit x -> - ∞ [1 / e^(2x)] = 0)
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2007-10-18 17:18:21 · answer #1 · answered by Madhukar 7 · 0⤊ 1⤋
The function that you wrote is the definition for the hyperbolic tangent, y=tanh(x). The limit is -1.
2007-10-18 17:18:30 · answer #2 · answered by The Prince 6 · 0⤊ 0⤋
multiply numerator and denominator by e^(-x).
Then you get
lim x->infinity 1-e^(-2x) / 1+e^(-2x) = 1-0/1+0 = 1
2007-10-18 17:16:17 · answer #3 · answered by q_midori 4 · 0⤊ 1⤋
Answer is -1
infinity/infinity=indefinite
use derivative rule
divide each term by e^x
(1-e^(-2x)/(1+e^(-2x))
use derivative (hospital rule)
2e^-2x/-2e^-2x=2
limit = -1
2007-10-18 17:20:59 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋