Evaluate this limit as x approaches infinity:
3/((e^-6x)-7)
2006-11-15 17:17:22 · 4 answers · asked by Mist 1 in Science & Mathematics ➔ Mathematics
Look at the denominator: e^-6x - 7; as x gets very large, e^-6x gets very small and soon becomes insignificant with respect to -7, so the limit will be -3/7.
2006-11-15 17:20:43 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
at positive infinity the term e^-6x is becoming zero so you get 3/(0-7) or just -3/7.
in the other direction, to negative infinity, e^-6x is becoming very very large, so the fration is going to get closer and closer to 0.
2006-11-15 17:21:31 · answer #2 · answered by xian gaon 2 · 0⤊ 0⤋
as x---> infinity, s^-6x---> 0
3/((e^-6x)-7) becomes 3/(-7)=-3/7
2006-11-15 17:20:22 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
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2006-11-15 17:23:28 · answer #4 · answered by satyagrahi 2 · 0⤊ 0⤋