# What are the asymptote(s) and hole(s), if any, of  f(x) =sin(pix)/x?

Dec 16, 2017

Hole at $x = 0$ and a horizontal asymptote with $y = 0$

#### Explanation:

First you have to calculate the zero marks of the denominator which in this case is $x$ therefore there's a vertical asymptote or a hole at $x = 0$. We aren't sure whether this is a hole or asymptote so we have to calculate the zero marks of the numerator
$\iff \sin \left(\pi x\right) = 0$
$\iff \pi x = 0 \mathmr{and} \pi x = \pi$
$\iff x = 0 \mathmr{and} x = 1$
As you see we have a common zero mark. This means that it's not an asymptote but a hole (with $x = 0$) and because $x = 0$ was the only zero mark of the denominator that means that they're are no vertical asymptotes.

Now we take the $x$-value with highest exponent of the denominator and of the numerator and divide them by each other.
but because there is only one kind of exponent of $x$ , the function $f \left(x\right)$ doesn't change.
$\iff \sin \frac{\pi x}{x}$
Now, if the exponent is bigger in the numerator than of the denominator that means that there's a diagonal or a curved asymptote. Otherwise, there's a straight line. In this case, it's going to be a straight line. Now you divide the a values of the numerator by the a value of the denominator.
$\iff S \in \frac{\pi}{1}$
$\iff \frac{0}{1}$
$\iff 0$
$\iff y = 0$ $=$ the horizontal asymptote