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
Thanks
On
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.
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.