0
$\begingroup$

I'm teaching a differential calculus course and incorrectly taught my students that to find oblique asymptotes you multiply and divide the fraction by the reciprocal of the largest power of x in the denominator, and what is left after taking the limit to infinity is the oblique asymptote.

A student did some exploring, discovered it was wrong, and informed me. Since then I've been curious why you must perform polynomial long division to find the oblique asymptotes.

Everything I can find online about it merely states what you have to do, not why you must perform long division.

To me it seems what I originally thought makes sense, although it is wrong. Why is it wrong and the correct way is via polynomial long division?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

You don't "need" to do long division; but that is a quick way to try to figure it out when you are dealing with rational functions.

Essentially, an oblique asymptote is a line $y=ax+b$, $a\neq 0$, with the property that $\lim_{x\to\infty}(f(x)-(ax+b)) = 0$. This is the same as verifying that $\lim_{x\to\infty}(f(x)-ax) = b$.

Now, the issue is that you don't know what the value of $a$ is ahead of time. How can we find a possible value of $a$? Well, if $f(x)$ is approaching $ax+b$, then $\frac{f(x)}{x}$ is approaching $a + \frac{b}{x}$, which goes to $a$ as $x\to\infty$. That is:

$\lim_{x\to\infty}(f(x)-ax)$ exists only if $\lim_{x\to\infty}\frac{f(x)}{x} = a$.

Proof. If $\lim_{x\to\infty}(f(x)-ax) = b$, then $\lim_{x\to\infty}(f(x)-(ax+b))=0$, so $\lim_{x\to\infty}(\frac{f(x)}{x} - a - \frac{b}{x}) = 0$.

Since $\lim_{x\to\infty}(a+\frac{b}{x}) = a$, the limit of the difference exists if and only if $\lim_{x\to\infty}\frac{f(x)}{x}$ exists, and equals $a$. $\Box$

So: we can first test to see whether $\lim_{x\to\infty}\frac{f(x)}{x}$ exists; if it does, and equals $a$, then we can then check to see if $\lim_{x\to\infty}(f(x)-ax)$ exists; if it does, and equals $b$, then we conclude that $\lim_{x\to\infty}(f(x) - (ax+b)) = 0$, so $y=ax+b$ is an asymptote to $f(x)$. If either of those limits do not exist, then there is no oblique asymptote.

Note: this works for every kind of function, not just rational ones.

Now, in light of that, what happens with a rational function? Well, if $f(x)=\frac{p(x)}{q(x)}$, you first try to do $\lim_{x\to\infty}\frac{p(x)}{xq(x)}$ and see if it exists; if it does and equals $a$, then you check $\lim_{x\to\infty}(\frac{p(x)}{q(x)} - ax)$ and see if it exists; and if so you get the asymptote.

But you can do a shortcut: $\lim_{x\to\infty}\frac{p(x)}{xq(x)}$ exists if and only if $\deg(xq(x))\geq \deg(p(x))$, which holds if and only if $\deg(q(x))+1\geq \deg(p(x))$. Since the cases with $\deg(q)\geq \deg(p)$ are already familiar (they yield horizontal asymptotes), we can restrict to the case $\deg(q)=\deg(p)-1$. But in that case, doing long division we can express $\frac{p(x)}{q(x)}$ as $ax+b + \frac{r(x)}{q(x)}$ (with $r(x)$ a polynomial of degree strictly smaller than $q(x)$, so that the fraction goes to $0$ as $x\to\infty$), and in that case it is immediately clear that $f(x)$ approaches $ax+b$ as $x\to\infty$. That is: you don't actually have to do limits in this case, you just have to do algebra.

What you propose, to divide both by the highest power of $x$ in the denominator, does not determine both $a$ and $b$. In fact, that only detects horizontal asymptotes. Since you are multiplying by $1$, you are just calculating $\lim_{x\to\infty} f(x)$. This exists if and only if $f(x)$ approaches a horizontal line $y=b$ (that is, $a=0$).

Obviously, the same arguments work for $x\to-\infty$.

$\endgroup$
1
  • $\begingroup$ Thank, you, yes I see it now, you are only able to distribute the limit across the division if the limit exists as a finite number, which only happens if the rational function has a horizontal asymptote. $\endgroup$
    – Aphyd
    Commented Jan 28, 2020 at 23:15
0
$\begingroup$

Not necessary to perform long division as it is not clear why it should give slant asymptote any way. Better to go like this below, If y= mx+c is asymptote then it must be true that lim x tends to infinity of f(x)-(mx+c) is zero. Once it is true (understood). Find limit as x tends to infinity (f(x)-mx-c)/(x) which any way has to be zero because numerator tends to zero and denominator tends to infinity. This limit should give us the slope m. Next the limit as x tends to infinity of f(x)-mx will give us the c. Why?, the limit f(x)-mx-c+c is same as f(x)-mx. and the limit f(x)-mx-c should be already zero. So this limit gives us c. When we actually plot the graphs of f(x) and mx+c we can see it is really an asymptote.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .