[Limit] Algebraic Manipulation of Rational Function

[TadeusPrastowo]

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:
<br /> \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}}<br />

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:
<br /> \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)}<br />
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.

 

[Mark44]

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:
<br /> \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}}<br />

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:
<br /> \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)}<br />
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.

 

[TadeusPrastowo]

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.

 

[TadeusPrastowo]

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.

 

[Mark44]

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.

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.

 

[TadeusPrastowo]

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).

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.