# How do I find the vertical asymptotes of f(x) = tanπx?

Sep 7, 2015

There are infinite (countable) number of asymptotes described by the following expression for $x$:
$x = \frac{1}{2} + N$, where $N$ - any integer number.

#### Explanation:

By definition, the vertical asymptote of a function is a vertical line on the coordinate plane that intersects the X-axis at a point where the value of a function is undefined and is infinitely increasing to $+ \infty$ or infinitely decreasing to $- \infty$ as its argument $x$ approaches to this point.

Since the definition of a function $\tan \left(\phi\right)$ is $\sin \frac{\phi}{\cos} \left(\phi\right)$, function $\tan \left(\right)$ has asymptotes wherever $\cos \left(\right)$ is equal to zero.

So, to get an asymptotes of function $\tan \left(\pi x\right)$, we have to resolve an equation
$\cos \left(\pi x\right) = 0$

As we know, function $\cos \left(\phi\right)$ represents an abscissa (X-coordinate) of a point on a unit circle that is an endpoint of a vector at an angle $\phi$ with a positive direction of the X-axis. So, it equals to zero when this vector is either vertically directed up or down along the Y-axis, which corresponds to angles $\frac{\pi}{2}$ and $- \frac{\pi}{2}$. Adding periodicity, we can say that an angle must be equal to $\frac{\pi}{2} + \pi N$, where $N$ - any integer number.

At $N = 0$ we get $x = \frac{\pi}{2}$.
At $N = - 1$ we get $x = - \frac{\pi}{2}$.
At $N = 1$ we get $x = 3 \frac{\pi}{2}$ (same angle as $- \frac{\pi}{2}$).
At $N = - 2$ we get $x = - 3 \frac{\pi}{2}$ (same angle as $\frac{\pi}{2}$).
etc.

So, we have a solution for our equation:
$\pi x = \frac{\pi}{2} + \pi N$ or
$x = \frac{1}{2} + N$,
where $N$ - any integer number.

These values of $x$ are those points where function $\tan \left(\pi x\right)$ has vertical asymptotes.