Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.

Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.

I would like to find out, though, what is wrong with the following incorrect way of finding the asymptote:

$\displaystyle y=\frac{x^2+x-6}{x+2}=\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}\approx\frac{x+1}{1}=x+1$, so $y=x+1$ is the slanted asymptote.


4 Answers 4


The trouble is that the numerator in $$\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}$$ stays large, so a small error in the denominator is magnified. Better to write $$ \frac{1}{1+\frac{2}{x}} = 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots $$ and multiply; the infinite series indicated converges for $|x| > 2.$

$$ \left( x+1-\frac{6}{x} \right) \left( 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots \right) = x - 1 - \frac{4}{x} + \frac{8}{x^2} + \cdots $$

  • 1
    $\begingroup$ To be precise, since the numerator is on the order of $x$, an error on the order of $1/x$ in the denominator creates an overall error on the order of $x(1/x)=1$ in the quotient, which fails to go to zero. $\endgroup$
    – Ian
    Oct 8, 2016 at 18:37
  • $\begingroup$ @Ian I put in just a little more. $\endgroup$
    – Will Jagy
    Oct 8, 2016 at 18:50

By definition, the function $f(x)$ has as an oblique (slanted) asymptote the straight line $y=\alpha x+\beta$ if: $$ \lim_{x\to+\infty}\big(f(x)-(\alpha x+\beta)\big)=0 $$ or if: $$ \lim_{x\to-\infty}\big(f(x)-(\alpha x+\beta)\big)=0 $$ It easy to check that while your result $y=x-1$ satisfies the above definition, the line $y=x+1$ does not, since: $$ \lim_{x\to\pm\infty}\big(f(x)-(x+1)\big)=-2 $$ P.S.: What you have actually shown via your second argument is that: $$ \lim_{x\to\pm\infty}\frac{f(x)}{x+1}=1 $$ which is however irrelevant to the notion of oblique (slanted) asymptote.


The problem with your second approach is that you've kept more precision than your approximation actually has — you can say $y \approx x$, but you don't have enough precision to clarify that more specifically to $y \approx x+1$ (or any other translate).

In more detail,

$$ \frac{1}{1 + \frac{2}{x}} \approx 1 - \frac{2}{x} $$

and consequently,

$$ \frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}} \approx \left( x+1-\frac{6}{x} \right) \left(1 - \frac{2}{x} \right) \approx x \cdot 1 + 1 \cdot 1 - x \cdot \frac{2}{x}$$

By neglecting the $\frac{2}{x}$ term of the denominator, you neglect the $x \cdot \frac{2}{x}$ term of this approximation — but that term is $-2$, so you're neglecting a nonnegligible quantity!

Keeping the $\frac{2}{x}$ term around, the above approximation gives $x-1$, as desired.

For more rigor, you can use big O notation:

$$\frac{1}{1 + \frac{2}{x}} = 1 - \frac{2}{x} + O(x^{-2}) $$ $$ \frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}} = \left( x+1+O(x^{-1}) \right) \left(1 - \frac{2}{x} + O(x^{-2}) \right) = x - 1 + O(x^{-1}) $$


$\frac{y}{x+1} \;\to\; 1$ indeed, but:

$$y - (x+1) = \frac{x^2 + x - 6 - x^2 - 3 x - 3}{x+2} = \frac{-2 x - 9}{x+2} \quad \to \quad -2$$

which is why the slope of $x+1$ is correct, but its intercept is off by $-2$.


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