annoyingly simple problem - rational functions and limits at infinity


by damian6961
Tags: annoyingly, functions, infinity, limits, rational, simple
damian6961
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#1
Apr5-09, 01:23 PM
P: 1
Hi all

This is my first post so please be gentle with me!

Limit of this rational function as x approaches infinity?

f(x) = (x^3 - 2x)/(2x^2 - 10)

I was under the impression that if the degree of the polynomial of the numerator exceed that of the denominator then there could be no horizontal asymptote. Is this correct?

I've used l'hopitals rule and found the limit to be 3x/2. I've been told the limit as x tends to infinity is x/2. Which is the correct solution and why? This has been driving me crazy!!

Damian
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CRGreathouse
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#2
Apr5-09, 01:50 PM
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It's x/2. Just divide everything by 2x^2.
elect_eng
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#3
Apr5-09, 01:53 PM
P: 370
Or, just retain the highest order term in the numerator and also in the denominator.

This leaves a ratio of [tex] {{x}\over{2}}[/tex]

adriank
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#4
Apr7-09, 01:12 AM
P: 534

annoyingly simple problem - rational functions and limits at infinity


The limit is [tex]\infty[/tex]. You can find that by taking the above suggestions and taking x/2 as x goes to infinity, but x/2 itself is not the limit.
elect_eng
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#5
Apr7-09, 09:23 AM
P: 370
Quote Quote by adriank View Post
The limit is [tex]\infty[/tex]. You can find that by taking the above suggestions and taking x/2 as x goes to infinity, but x/2 itself is not the limit.
You misunderstand what I'm saying. The function approaches a "limiting" (asymptotic) form of x/2 in the limit at x goes to infinity. It doesn't matter if the value of the function goes to infinity as x goes to infinity. Electrical engineers use this trick all the time to find out the high frequency response of a system. We always have frequency response as a ratio of polynomials in frequency. We just keep the highest order terms in the numerator and in the denominator to find the high frequency response.
adriank
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#6
Apr8-09, 12:15 AM
P: 534
I'm not misunderstanding anything. The question didn't ask about the asymptotic behaviour of the function as x goes to infinity; it only asked about the limit. I was giving a clear answer to the question.
elect_eng
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#7
Apr8-09, 08:32 AM
P: 370
Quote Quote by adriank View Post
I'm not misunderstanding anything. The question didn't ask about the asymptotic behaviour of the function as x goes to infinity; it only asked about the limit. I was giving a clear answer to the question.
Well, I read the question differently. It says "what is the limit of the rational function?". This implies finding the limiting functional form which is born out by the answer. He tried to solve the problem as if he needs the numerical limit, which is why he is confused.

If you are saying that the wording of the question is vague and that the answer of infinity could be acceptable as an answer, I won't argue about that.
adriank
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#8
Apr8-09, 08:09 PM
P: 534
I didn't find the problem statement unclear or ambiguous at all. Oh well.
AUMathTutor
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#9
Apr8-09, 09:09 PM
P: 490
The limit of the function as x approaches infinity is undefined (it tends towards infinity).

The f(x) - x/2 goes to zero in the limit of large n, implying that the function f(x) ~ x/2 asymptotically.
elect_eng
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#10
Apr9-09, 11:24 AM
P: 370
Quote Quote by adriank View Post
I didn't find the problem statement unclear or ambiguous at all. Oh well.
Yes, I see your point. Math requires precise statements. I shouldn't read things into it.
CRGreathouse
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#11
Apr9-09, 11:55 AM
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Quote Quote by AUMathTutor View Post
The limit of the function as x approaches infinity is undefined (it tends towards infinity).

The f(x) - x/2 goes to zero in the limit of large n, implying that the function f(x) ~ x/2 asymptotically.
I always use extended reals for limits, so I would say
[tex]\lim_{x\to\infty}f(x)=+\infty.[/tex]

If you really want to get technical, what the original poster was asking for is the first term of the asymptotic expansion of f(x) about infinity:
[tex]f(x)=\frac x2+\frac{3}{2x}+\frac{15}{2x^3}+\frac{75}{2x^5}+\cdots[/tex]


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