Horizontal asymptote of a parametric function

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  • #1
greg_rack
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Homework Statement:
Consider ##f(x)=\frac{(a-2)x^3+x^2}{ax^2+6x+1}##:
for which values of ##a## has it an horizontal asymptote?
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I'll write my procedure:
$$\lim_{x\to\infty}[\frac{(a-2)x^3+x^2}{ax^2+6x+1}]\rightarrow\frac{x(a-2)}{a}\in \mathbb{R}$$
And now, assumed that everything's correct, how do I assign ##a## a value for having that limit finite and ##\in \mathbb{R}##, and so an horizontal asymptote?
 

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  • #2
Office_Shredder
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Can you describe in words why for at least most values of a you don't get a horizontal asymptote? For example what if a=1?
 
  • #3
greg_rack
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Can you describe in words why for at least most values of a you don't get a horizontal asymptote? For example what if a=1?
For most ##a## values, the limit would be infinite, and of course for a horizontal asymptote we want the limit to have a finite value
 
  • #4
Office_Shredder
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Yep. In order to keep the asymptote from being a line that goes off to infinity, you need to prevent the numerator from having a higher degree in x than the numerator. What value of a gives you that?
 
  • #5
greg_rack
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Yep. In order to keep the asymptote from being a line that goes off to infinity, you need to prevent the numerator from having a higher degree in x than the numerator. What value of a gives you that?
Well right, for ##a=2## the numerator is ##0## and the asymptote is ##y=0##!
But how could we generalize this in saying that "you need to prevent the numerator from having a higher degree in x than the denominator"? Is it always true...?
 
  • #6
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But how could we generalize this in saying that "you need to prevent the numerator from having a higher degree in x than the denominator"? Is it always true...?
For a rational function (the quotient of two polynomials), if the degree of the numerator is equal to the degree of the denominator, there will be a horizontal asymptote.

Further, if ##f(x) = \frac{a_nx^n + \text{ lower degree terms}}{b_nx^n + \text{ lower degree terms}}##, then the equation of the hor. asymptote is ##y = \frac{a_n}{b_n}##.
 
  • #7
pasmith
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Well right, for ##a=2## the numerator is ##0## and the asymptote is ##y=0##!

Try again. You've set the leading term in the numerator to zero, but there is another term there.
 
  • #8
greg_rack
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For a rational function (the quotient of two polynomials), if the degree of the numerator is equal to the degree of the denominator, there will be a horizontal asymptote.

Further, if ##f(x) = \frac{a_nx^n + \text{ lower degree terms}}{b_nx^n + \text{ lower degree terms}}##, then the equation of the hor. asymptote is ##y = \frac{a_n}{b_n}##.
Got it, so in order to avoid infinite limits, for rational functions, the numerator must have an equal or lower degree than the denominator: ##y=\frac{a}{\infty}## or ##y=\frac{a}{b}##
 
  • #9
greg_rack
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Try again. You've set the leading term in the numerator to zero, but there is another term there.
Yes, I have even noticed that with ##a=2## I'll end up with the indeterminate form ##0\cdot\infty##... how could I solve this then?
 
  • #10
etotheipi
I'll write my procedure:
$$\lim_{x\to\infty}[\frac{(a-2)x^3+x^2}{ax^2+6x+1}]\rightarrow\frac{x(a-2)}{a}\in \mathbb{R}$$

[Not that this is relevant to the question, but I think you did the long division wrong, i.e. you missed an additive constant. Should be$$\lim_{x \to \infty} \left( \frac{(a-2)x^3 + x^2}{ax^2 + 6x + 1} \right) = \frac{a-2}{a}x + \frac{12-5a}{a^2}$$this would be noticeable for small ##a##, perhaps less so for large ##a##.]
 
  • #11
greg_rack
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[Not that this is relevant to the question, but I think you did the long division wrong, i.e. you missed an additive constant. Should be$$\lim_{x \to \infty} \left( \frac{(a-2)x^3 + x^2}{ax^2 + 6x + 1} \right) = \frac{a-2}{a}x + \frac{12-5a}{a^2}$$this would be noticeable for small ##a##, perhaps less so for large ##a##.]
I cannot understand where does the additive constant comes from. The ##\frac{a-2}{a}x## is for the fact that the limit of a rational function is equal to the limit of the ratio of its highest degree terms
 
  • #12
etotheipi
I cannot understand where does the additive constant comes from. The ##\frac{a-2}{a}x## is for the fact that the limit of a rational function is equal to the limit of the ratio of its highest degree terms

Nah, that's only when the degrees are equal, i.e. as @Mark44 pointed out, if$$y = \frac{a_1 x^n + \dots}{a_2 x^n + \dots}$$then the asymptote is ##y = a_1/a_2##. If in a different instance the degree of the denominator greater than that of the numerator, then the horizontal asymptote is ##y=0##.

But if the degree of the numerator is greater than that of the denominator, you need to do a long division [or "filling in the gaps"] to figure out the slant asymptote. In this case,$$(a-2)x^3 + x^2 = (ax^2 + 6x + 1) \left(\frac{a-2}{a}x + \frac{12-5a}{a^2} \right) + \left( \frac{-a^2 +32a - 72}{a^2}x - \frac{12 - 5a}{a^2} \right)$$ $$\frac{(a-2)x^3 + x^2}{ax^2 + 6x + 1} =\left(\frac{a-2}{a}x + \frac{12-5a}{a^2} \right) + \frac{1}{ax^2 + 6x + 1}\left( \frac{-a^2 +32a - 72}{a^2}x - \frac{12 - 5a}{a^2} \right)$$In limit ##x \rightarrow \infty##, for instance, that second term goes to zero and you get$$\lim_{x \to \infty} \left( \frac{(a-2)x^3 + x^2}{ax^2 + 6x + 1} \right) = \frac{a-2}{a}x + \frac{12-5a}{a^2}$$If you try plotting it on some graphing software, you'll see the vertical shift.
 
  • #13
greg_rack
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Nah, that's only when the degrees are equal, i.e. as @Mark44 pointed out, if$$y = \frac{a_1 x^n + \dots}{a_2 x^n + \dots}$$then the asymptote is ##y = a_1/a_2##. If in a different instance the degree of the denominator greater than that of the numerator, then the horizontal asymptote is ##y=0##.

But if the degree of the numerator is greater than that of the denominator, you need to do a long division [or "filling in the gaps"] to figure out the slant asymptote. In this case,$$(a-2)x^3 + x^2 = (ax^2 + 6x + 1) \left(\frac{a-2}{a}x + \frac{12-5a}{a^2} \right) + \left( \frac{-a^2 +32a - 72}{a^2}x - \frac{12 - 5a}{a^2} \right)$$ $$\frac{(a-2)x^3 + x^2}{ax^2 + 6x + 1} =\left(\frac{a-2}{a}x + \frac{12-5a}{a^2} \right) + \frac{1}{ax^2 + 6x + 1}\left( \frac{-a^2 +32a - 72}{a^2}x - \frac{12 - 5a}{a^2} \right)$$In limit ##x \rightarrow \infty##, for instance, that second term goes to zero and you get$$\lim_{x \to \infty} \left( \frac{(a-2)x^3 + x^2}{ax^2 + 6x + 1} \right) = \frac{a-2}{a}x + \frac{12-5a}{a^2}$$If you try plotting it on some graphing software, you'll see the vertical shift.
I'll quote what my textbook says about the limit of a rational function(it's in Italian so there may be a few language errors/imprecisions):
"To calculate the limit of a ratio of two polynomials for ##x\rightarrow +-\infty## is enough to calculate the limit of the ratio for its terms of maximum degree"
And that is easily demonstrable by collecting(I'm not sure it's the right word) the maximum degree terms since inside the parenthesis everything will be ##0##(since it's the limit tending to ##\infty##).
 
  • #14
etotheipi
I would get a different textbook 😜. Unless it's referring to rational functions where the degrees of the numerator and denominator are equal.
 
  • #15
greg_rack
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I would get a different textbook 😜. Unless it's referring to rational functions where the degrees of the numerator and denominator are equal.
IMG_7668.jpg
Have a look! It looks legit to me hahaha
 
  • #16
etotheipi
Well this textbook is talking about something different. Notice that
$$x^3 + 5x^2 + 6x - 1 = (x^2 + 7x+1)(x-2) + (19x+1)$$ $$\frac{x^3 + 5x^2 + 6x - 1}{x^2 + 7x + 1} = x-2 + \frac{19x+1}{x^2 + 7x + 1}$$Hence,$$\lim_{x\to\infty} \frac{x^3 + 5x^2 + 6x - 1}{x^2 + 7x + 1} = x-2$$And notice the additive constant ##-2## is present. Try graphing the function!

##x## and ##x-2## both tend toward each other as ##x\rightarrow \infty##, so it doesn't matter for the purposes of the textbook, which is just working out a limit. But if you want to find the asymptote, you can't do it by dividing leading coefficients!
 
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  • #17
pasmith
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Yes, I have even noticed that with ##a=2## I'll end up with the indeterminate form ##0\cdot\infty##... how could I solve this then?

If [itex]a = 2[/itex] then [tex]
\frac{(a - 2)x^3 + x^2}{ax^2 + 6x + 1} = \frac{x^2}{2x^2 + 6x + 1}.[/tex] You can't ignore the [itex]x^2[/itex] term if the coefficient of the [itex]x^3[/itex] term is zero.
 
  • #18
greg_rack
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If [itex]a = 2[/itex] then [tex]
\frac{(a - 2)x^3 + x^2}{ax^2 + 6x + 1} = \frac{x^2}{2x^2 + 6x + 1}.[/tex] You can't ignore the [itex]x^2[/itex] term if the coefficient of the [itex]x^3[/itex] term is zero.
So now, with ##a=2## the limit of ##\frac{x^2}{2x^2 + 6x + 1}## is finite and is ##y=\frac{x^2}{2x^2}##... or isn't it?
 
  • #19
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View attachment 271136
Have a look! It looks legit to me hahaha
In the page from your book, they're just talking about the limit, and not about a horizontal or slant asymptote.
So now, with ##a=2## the limit of ##\frac{x^2}{2x^2 + 6x + 1}## is finite and is ##y=\frac{x^2}{2x^2}##... or isn't it?
Yes, the limit, as ##x \to \infty##, is finite, and is equal to 1/2.
$$\lim_{x\to \infty}\frac {x^2}{2x^2 + 6x + 1} = \lim_{x\to \infty}\frac {x^2(1)}{x^2(2 + 6/x + 1/x^2)} = \frac 1 2$$
 

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