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For f(x) = (2x + 5) / |3x - 4|, use graphs and tables to find the limit as x approaches infinity of f(x) and the limit as x approaches negative infinity of f(x)... Also identify any horizontal asymptotes...
When i graph the function in my graphing calculator, it looks like y will never reach a specific value as x approaches pos or neg infinity.. i went to table and checked at x value or 20000000 and -200000000 and i get 2/3 and -2/3 respectively
is this correct?
limit as x approaches pos inf = 2/3
limit as x approaches neg inf = -2/3
I'm not sure about the horizontal asymptotes?
would they be y = 2/3 and y = -2/3? I think? or is it something in between that?
Thanks for the input.
Horizontal asymptotes are the limits as x approaches infinity.
If
where c is a constant, then the horizontal asymptote is x = c. Same applies for negative infinity.You're answers are correct. Whenever you have a polynomial division, where a*x^n and b*x^n are the highest terms for each the numerator and the denominator, than a horizontal asymptote exists at a/b. Note that both n's have to be the same. In other words, this does not apply to 5x^3 / 2x^2.
Edited to add:
if you have a*x^n and b*x^m as the highest terms in the polynomial division, then:
if n > m, the function goes to infinity
if n < m, the function has a horizontal asymptote at x = 0
if n = m, the function has a horizontal asymptote at a/b
Last edited by Ricky (2006-01-02 11:22:36)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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If x->+oo the function f is:
f(x)=(2x+5)/(3x-4). We'll find the limit:
Last edited by krassi_holmz (2006-01-02 11:30:05)
IPBLE: Increasing Performance By Lowering Expectations.
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Analogic,
2
lim f(x) = - ---
x--> -oo 3
IPBLE: Increasing Performance By Lowering Expectations.
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