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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.

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    $\begingroup$ $y=1$ 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
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    $\begingroup$ Yes, that's right $\endgroup$ – Vasya Mar 28 at 16:19
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Yup. Horizontal asymptotes are happy to be intersected as many times as you like.

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  • $\begingroup$ 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

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