Using limits to find the asymptote of a function $y=f(x)$ is usually done with limits as : if the asymptote is of the form $y=mx+c$ then :

$m=\lim\limits_{x\to\infty} \dfrac{f(x)}{x}$ $\space$and$\space$ $c=\lim\limits_{x\to\infty} (f(x)-mx)$

Using the same formula can we also find the asymptote of $\space \space y=\tan(x)$ ?


I'm personally having problem while solving for m...please help


  • $\begingroup$ The formula is valid only for asymtotes of finite slope for $x \rightarrow \infty$. $\endgroup$ Commented Apr 20, 2015 at 16:31

2 Answers 2


The formula you propose is only valid for asymptotes of the form $$y = mx + q,$$ where $m \in \mathbb R \backslash \{0\}$ and $q \in \mathbb R$. Since $m \in \mathbb R$, these lines cannot be vertical.

On the other hand, the asymptotes you seek are of the form $x = q$. You can find them by researching all the points $\alpha$ for which $$\left|\lim_{x \to \alpha^-} f(x)\right| = +\infty\qquad\text{or}\qquad\left|\lim_{x \to \alpha^+} f(x)\right| = + \infty.$$ This usually happens where the function does not exist, but it isn't always the case.


In addition to what others replied , consider its inverse function

$$ y = f(x) = \tan^{-1} x $$

$$\left|\lim_{x \to +\infty} f(x)\right| = \pi/2 $$ which is a horizontal asymptote.

When you switch back to the original function $ x = \pi/2$ is now a vertical asymptote.

$$\left|\lim_{x \to \pi/2 } f(x)\right| = +\infty, $$ approaching from left.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .