# What are the asymptote(s) and hole(s), if any, of  f(x) =cos((pix)/2)/((x-1)(x+2))?

Dec 8, 2017

$f \left(x\right)$ has a hole at $x = 1$, a vertical asymptote at $x = - 2$ and a horizontal asymptote $y = 0$ (according to modern usage).

#### Explanation:

Given:

$f \left(x\right) = \cos \frac{\frac{\pi x}{2}}{\left(x - 1\right) \left(x + 2\right)}$

Note that the denominator is zero when $x = 1$ or $x = - 2$.

As $x \to 1$, then $\cos \left(\frac{\pi x}{2}\right) \to 0$. With both numerator and denominator tending to $0$, we may have a hole or we may have an asymptote.

Substituting $t = x - 1$, we have:

${\lim}_{x \to 1} f \left(x\right) = {\lim}_{t \to 0} \cos \frac{\frac{\pi t}{2} + \frac{\pi}{2}}{t \left(t + 3\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} f \left(x\right)} = {\lim}_{t \to 0} - \sin \frac{\frac{\pi t}{2}}{\frac{\pi t}{2}} \cdot \left(\frac{\frac{\pi}{2}}{t + 3}\right)$

$\textcolor{w h i t e}{{\lim}_{x \to 1} f \left(x\right)} = - 1 \cdot \frac{\pi}{6}$

$\textcolor{w h i t e}{{\lim}_{x \to 1} f \left(x\right)} = - \frac{\pi}{6}$

So $f \left(x\right)$ has a hole at $x = 1$

As $x \to - 2$, then $\cos \left(\frac{\pi x}{2}\right) \to \cos \left(- \pi\right) = - 1$

So the numerator is non-zero, while the denominator is zero. So there is a vertical asymptote at $x = - 2$

For any real value of $x$, we have $\left\mid \cos \left(\frac{\pi x}{2}\right) \right\mid \le 1$, while $\left(x - 1\right) \left(x + 2\right) \to \infty$ as $x \to \pm \infty$.

Hence $f \left(x\right)$ has a horizontal asymptote $y = 0$.

Note that historically some people would not count this as an asymptote since the graph of $f \left(x\right)$ crosses it infinitely many times, but it does actually tend towards the asymptote.

graph{cos((pix)/2)/((x-1)(x+2)) [-10.42, 9.58, -1.2, 1.2]}