e^3x - e^-3x
lim __________
x-> ∞ e^3x + e^-3x
how to solve this?
please show steps leading to answer!
lim = 1 , as x-> ∞
Well lets think about it.
e^-3x is the same thing as 1/e^3x
1/e^3x is gonna go to 0 when we take the limit.
So we will be left with whatever:
e^3x / e^3x does
And those are the same terms divided by each other so itll be 1.
(e^(3x) - e^(-3x)) / (e^(3x) + e^(-3x)) =>
((e^(6x) - 1) / e^(3x)) / ((e^(6x) + 1) / e^(3x)) =>
(e^(6x) - 1) / (e^(6x) + 1) =>
(e^(6x) + 1 - 2) / (e^(6x) + 1) =>
(e^(6x) + 1) / (e^(6x) + 1) - 2 / (e^(6x) + 1) =>
1 - 2 / (1 + e^(6x))
x goes to infinity
1 - 2 / (1 + e^(6 * inf)) =>
1 - 2 / (1 + e^(inf)) =>
1 - 2 / (1 + inf) =>
1 - 2 / (inf) =>
1 - 0 =>
1
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Verified answer
lim = 1 , as x-> ∞
Well lets think about it.
e^-3x is the same thing as 1/e^3x
1/e^3x is gonna go to 0 when we take the limit.
So we will be left with whatever:
e^3x / e^3x does
And those are the same terms divided by each other so itll be 1.
(e^(3x) - e^(-3x)) / (e^(3x) + e^(-3x)) =>
((e^(6x) - 1) / e^(3x)) / ((e^(6x) + 1) / e^(3x)) =>
(e^(6x) - 1) / (e^(6x) + 1) =>
(e^(6x) + 1 - 2) / (e^(6x) + 1) =>
(e^(6x) + 1) / (e^(6x) + 1) - 2 / (e^(6x) + 1) =>
1 - 2 / (1 + e^(6x))
x goes to infinity
1 - 2 / (1 + e^(6 * inf)) =>
1 - 2 / (1 + e^(inf)) =>
1 - 2 / (1 + inf) =>
1 - 2 / (inf) =>
1 - 0 =>
1