[Limit] Algebraic Manipulation of Rational Function

In summary, the conversation discusses the process of finding an oblique asymptote for a rational function as x grows without bound. The rule of thumb is to divide by the highest power in the denominator, and if the degree of the numerator is one more than the degree of the denominator, then there will be an oblique asymptote. To find it, polynomial long division can be used to write the function as a linear part and a proper rational function.
  • #1
TadeusPrastowo
21
0
My https://www.amazon.com/dp/0073532320/?tag=pfamazon01-20 gives a rule of thumb to divide by the highest power in the denominator for the following problem to demonstrate a slant (oblique) asymptote:
[tex]
\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{4x+\frac{5}{x^2}}{-6-\frac{7}{x}}
[/tex]

But, the textbook gives no explanation whatsoever why it has to be the highest power in the denominator. As an alternative, I can actually do the following instead:
[tex]
\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{x^3\left(4+\frac{5}{x^3}\right)}{-x^2\left(6+\frac{7}{x}\right)}
[/tex]
However, my alternative gives me ##\lim_{x\to\infty} -\frac{2}{3}x##. That is, a linear equation of ##-\frac{2}{3}x##. The textbook, however, resorts to performing a long division of ##4x^3+5## by ##-6x^2-7x## to obtain ##\lim_{x\to\infty} \left(-\frac{2}{3}x+\frac{7}{9}+\frac{5+\frac{49}{9}x}{-6x^2-7x}\right)##. That is, a linear equation of ##-\frac{2}{3}x+\frac{7}{9}##.

I have checked numerically that the slant asymptote is indeed the textbook's ##-\frac{2}{3}x+\frac{7}{9}## instead of my alternative's ##-\frac{2}{3}x##.

My question is then, why theoretically my alternative is wrong?

Thank you very much.
 
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  • #2
TadeusPrastowo said:
My https://www.amazon.com/dp/0073532320/?tag=pfamazon01-20 gives a rule of thumb to divide by the highest power in the denominator for the following problem to demonstrate a slant (oblique) asymptote:
[tex]
\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{4x+\frac{5}{x^2}}{-6-\frac{7}{x}}
[/tex]

But, the textbook gives no explanation whatsoever why it has to be the highest power in the denominator. As an alternative, I can actually do the following instead:
[tex]
\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{x^3\left(4+\frac{5}{x^3}\right)}{-x^2\left(6+\frac{7}{x}\right)}
[/tex]
However, my alternative gives me ##\lim_{x\to\infty} -\frac{2}{3}x##. That is, a linear equation of ##-\frac{2}{3}x##. The textbook, however, resorts to performing a long division of ##4x^3+5## by ##-6x^2-7x## to obtain ##\lim_{x\to\infty} \left(-\frac{2}{3}x+\frac{7}{9}+\frac{5+\frac{49}{9}x}{-6x^2-7x}\right)##. That is, a linear equation of ##-\frac{2}{3}x+\frac{7}{9}##.

I have checked numerically that the slant asymptote is indeed the textbook's ##-\frac{2}{3}x+\frac{7}{9}## instead of my alternative's ##-\frac{2}{3}x##.

My question is then, why theoretically my alternative is wrong?

Thank you very much.
Finding the oblique asymptote is different from finding the limit as x gets very large. Your first limit says merely that, as x gets very large, your function approaches ##-\infty##. The last limit you showed wasn't complete. ##\lim_{x \to \infty} \frac{-2x}{3} = -\infty##.

What the book is doing is showing that as x gets very large, the rational function gets close to a particular straight line.
 
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  • #3
Mark44 said:
Finding the oblique asymptote is different from finding the limit as x gets very large.

I don't agree. In my opinion, analytically, as ##x## gets very large, the algebraic manipulation of the function definition must reveal the nature of the asymptote: whether it is horizontal or slanted.

As the numerical calculation shows that my alternative obtained by extracting the greatest exponents of both the numerator & denominator is not accurate, I want to know: why algebraically/analytically/theoretically is my alternative wrong (it is off by 7/9)?

Thank you very much.
 
  • #4
TadeusPrastowo said:
As the numerical calculation shows that my alternative obtained by extracting the greatest exponents of both the numerator & denominator is not accurate, I want to know: why algebraically/analytically/theoretically is my alternative wrong (it is off by 7/9)?

Specifically, I want to know the sound method to algebraically manipulate a rational function definition in finding the equation of the asymptote as ##x## grows without bound. For example:

1. Always start with a long division because of this reason.
2. Then, for the resulting terms, divide by the largest exponent in the denominator because of this reason.
3. So, it is clear that my alternative is unsound because of the above points show that my alternative is bla bla.

Thank you very much.
 
  • #5
TadeusPrastowo said:
However, my alternative gives me
##\lim_{x\to\infty} -\frac{2}{3}x##
. That is, a linear equation of ##-\frac{2}{3}x##.
No. As I said before, you haven't finished taking the limit. ##\lim_{x\to\infty} -\frac{2}{3}x = -\infty##. When you take a limit in which x is a variable, you don't end up with an expression that still involves x. That is your mistake.

TadeusPrastowo said:
But, the textbook gives no explanation whatsoever why it has to be the highest power in the denominator.
They're looking at the highest power of x that appears in either the numerator or denominator. For your problem, there's an x3 term in the numerator, and all other terms are of lower degree. By factoring out the the largest power of x in the denominator, they are left with an x term in the numerator, and all other terms in numerator or denominator go to either a constant or zero.

Again, taking the limit doesn't tell you anything about an oblique asymptote. It merely tells you the gross behavior of the rational function as x gets large or as x gets very negative.

If the degree of the numerator is one more than the degree of the denominator, then there will be an oblique asymptote. To find it, do polynomial long division to write the as a linear part and a proper rational function. IOW, as ##L(x) + \frac{p(x)}{q(x)}##. Here L(x) = ax + b for some constants a and b, and the degree of p < degree of q.
 
  • #6
Mark44 said:
No. As I said before, you haven't finished taking the limit. ##\lim_{x\to\infty} -\frac{2}{3}x = -\infty##. When you take a limit in which x is a variable, you don't end up with an expression that still involves x. That is your mistake. ... Again, taking the limit doesn't tell you anything about an oblique asymptote. It merely tells you the gross behavior of the rational function as x gets large or as x gets very negative.

Now I understand. Thank you very much for your patience. I took a look again at the formal definition of limit as ##x## grows unbounded and saw that indeed the requirement for ##\lim_{x\to \infty} f(x)## to exist is to have a horizontal asymptote, not a slant asymptote (i.e., a slant asymptote is a case similar to vertical asymptote where a limit does not exist).

Mark44 said:
If the degree of the numerator is one more than the degree of the denominator, then there will be an oblique asymptote. To find it, do polynomial long division to write the as a linear part and a proper rational function. IOW, as ##L(x) + \frac{p(x)}{q(x)}##. Here L(x) = ax + b for some constants a and b, and the degree of p < degree of q.

Thank you very much for pointing the obvious fact. I realize now that the textbook would like to point out that when the limit does not exist as ##x## grows without bound, performing a long polynomial division will show through either an oblique asymptote or a vertical asymptote.

Thank you very much for your answer & patience.
 

Related to [Limit] Algebraic Manipulation of Rational Function

What is a rational function?

A rational function is a mathematical expression with a polynomial in the numerator and denominator. It can be written as a ratio of two polynomial expressions, where both the numerator and denominator are not equal to zero.

What is the process for simplifying a rational function?

The first step is to factor both the numerator and denominator. Then, cancel out any common factors between the two. Finally, simplify the remaining terms, if possible.

How do I find the domain of a rational function?

The domain of a rational function is all the values of x that make the function defined. To find the domain, set the denominator equal to zero and solve for x. The domain will be all values of x except for the ones that make the denominator equal to zero.

What is a common mistake when manipulating rational functions?

A common mistake is forgetting to check for extraneous solutions. This can happen when simplifying or solving for the domain, as it may introduce solutions that do not make the original function defined.

How is solving a rational function different from solving a regular equation?

Solving a rational function involves finding values of x that make the function equal to a specific value, while solving a regular equation involves finding the value of x that makes the equation true. Also, when solving a rational function, it is important to check for extraneous solutions, as mentioned in the previous question.

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