# Why must you do polynomial long division to find oblique asymptotes of rational functions?

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?

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 $$x\to-\infty$$.

• 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. Commented Jan 28, 2020 at 23:15

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.