# Why do I have to perform polynomial division when trying to find slant asymptotes

When trying to find the slant asymptote of $$\frac{2x^2+x}{x-3}$$, the way I thought was correct is to divide everything by $$x$$ to get $$\frac{2x+1}{1-\frac{3}x}$$. All that was left was to say that as $$x$$ tends to $$\infty$$, $$\frac{3}x$$ tends to $$0$$, so the asymptote is $$2x+1$$. Spoiler: it was not. If I would do it the polynomial division way, I would get that the asymptote is $$2x+7$$. My question is: what is wrong with my way?

• I think you are mixing up limits with asymptotic behaviour. You did not prove $f(x)=\frac{2x^2+x}{x-3}\to 2x+1$ when $x\to\infty$ (i.e. better written: $f(x)-(2x+1)\to 0$ when $x\to\infty$): you have proven that $f(x)\sim 2x+1$ when $x\to\infty$ which is weaker. (After all, $2x+7\sim 2x+1$ itself!)
– user700480
Commented Jan 19, 2022 at 14:52
• Exactly what is the definition of "slant asymptote"? (Say $f$ is our function and $y=L(x)$ is the line; if the definition is $f(x)/L(x)\to1$ fine, but if the definition is $f(x)-L(x)\to0$ then you simply haven't shown that.) Commented Jan 19, 2022 at 14:56
• This might provide some insight. Commented Jan 19, 2022 at 15:01
• The distinction that @StinkingBishop is talking about is related to a confusing mismatch of terminology. We say that two functions $f$ and $g$ are asymptotic (at infinity) if $\lim_{x\to\infty}(f(x)/g(x))=1$. But we say that the non-vertical line $y=g(x)=mx+b$ is an asymptote (on the right) of the curve $y=f(x)$ if $\lim_{x\to\infty}(f(x)-g(x))=0$. (These both also have versions where $x\to-\infty$, of course.) And annoyingly, these are not the same! Commented Jan 19, 2022 at 15:45
• PS: I once asked a question on MathOverflow asking for the proper term for functions $f$ and $g$ such that $\lim_{x\to\infty}(f(x)-g(x))=0$, and nobody knew one. (The best advice seemed to be to just say ‘asymptotic’ with an ad-hoc adverb or a warning that the term is being used in a nonstandard sense.) Commented Jan 19, 2022 at 15:50

Asserting that $$2x+1$$ is an asymptote means that$$\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(2x+1)=0.$$But$$\frac{2x^2+x}{x-3}-(2x+1)=\frac{6x+3}{x-3}$$and therefore$$\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(2x+1)=6\ne0.$$On the other hand\begin{align}\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(ax+b)=0&\iff\lim_{x\to\infty}\frac{-a x^2+2 x^2+3 a x-b x+x+3 b}{x-3}=0\\&\iff\left\{\begin{array}{l}-a+2=0\\3a-b+1=0\end{array}\right.\\&\iff\left\{\begin{array}{l}a=2\\b=7.\end{array}\right.\end{align}

Other people have explained how you got the wrong answer, but another way that you can use your calculation as a step to a correct answer is to recognize that this method will give you the correct slope $$2$$ but will not tell you the constant term. So to find that, you take $$\lim _ { x \to \infty } \Big ( \frac { 2 x ^ 2 + x } { x - 3 } - 2 x \Big ) \text ,$$ and this will give you $$7$$. Therefore, the asymptote is $$2 x + 7$$.

Perhaps like this:

$$(2x+1)(1-3/x)^{-1};$$

Now for large $$x$$ expand the second factor:

$$(2x+1)\cdot$$

$$(1 +(-1)(-3/x)+O(1/x^2))=$$

$$2x+1+(6x)/x+O(1/x)=$$

$$2x+7+O(1/x);$$ and we are done.

Used: Binomial expansion

An option:

$$\dfrac{2x^2+x}{x-3}=$$

$$\dfrac{2((x-3)+3)^2+(x-3)+3}{x-3}=$$

$$2(x-3)+12+\dfrac{18} {x-3} +1+\dfrac{3}{x-3} =$$

$$2(x-3)+13+\dfrac{21}{x-3} =$$

$$2x+7+\dfrac{21}{x-3} .$$

Can you finish?