I'm having some difficulty understanding oblique asymptotes.

Let's have

$$f(x) = 2x^3+3x^2-12x$$

The oblique asymptote appears only when

$$\lim_{x\to \pm \infty} f(x) = \pm \infty$$

So I need to evaluate two limits.

$$\lim_{x\to + \infty} f(x) = +\infty$$


$$\lim_{x\to - \infty} f(x) = -\infty$$

Ok good, it seems we got two oblique asymptotes? One to the right and another to the left?

Let's focus on the one to the right. The slope is

$$m = \lim_{x\to +\infty} \frac{f(x)}{x}$$

Evaluating yields that $m = \infty$.

....... this can't be right. An infinite slope would be a straight vertical line.

What am I doing wrong?

  • $\begingroup$ You are wrongly interpreting the result. $x^3$ has an oblique asymptote and it's slope tends to infinity. No contradiction here. $\endgroup$ – uniquesolution Nov 10 '15 at 8:13

Your calculations are correct. An oblique occurs when a function tends toward a line as $x \to \pm \infty$. But in your case, the function grows much faster than a line, and there is no asymptote. (Note that the slope is not infinite, so there is no vertical line. Rather, the slope is steadily increasing.)

It would be different if you had, for instance, $f(x) = \dfrac{2x^3+3x^2-12x}{x^2+1}$, which is asymptotic to $2x+3$ as $x \to \infty$. Try graphing the function to see its behaviour.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.