Consider the following function: $f(x) = \frac{(x-2)^2}{(x^2-1)}$. To calculate the horizontal asymptote, we take the limit:

$\displaystyle{\lim_{x \to \infty}} f(x) = 1$

However, $f(x)$ still intersects with $y = 1$, namely at the point $(1.25;1)$.

So is $y = 1$ the horizontal asymptote? Because it still intersects with $y = 1$, but the function approaches 1 as x approaches to infinity.

  • 3
    $\begingroup$ $y=1$ is horizontal asymptote, you got it right. Consider function $y=\sin \frac{1}{x}$, it has infinitely many intersections with $y=0$ which is horizontal asymptote $\endgroup$ – Vasya Mar 28 at 16:08
  • $\begingroup$ Ah, that's really interesting. Thanks for the example! I assume vertical asymptotes can't be intersected, but horizontal asymptotes can right? $\endgroup$ – Stallmp Mar 28 at 16:14
  • 1
    $\begingroup$ Yes, that's right $\endgroup$ – Vasya Mar 28 at 16:19

Yup. Horizontal asymptotes are happy to be intersected as many times as you like.

  • $\begingroup$ Oh alright, I thought horizontal asymptotes are never intersected, so the two contradicting results confused me. I've learned that horizontal asymptotes are values of y which the function approaches, but will never intersect, so that lead to the confusion. $\endgroup$ – Stallmp Mar 28 at 16:07

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.