# What are the asymptote(s) and hole(s), if any, of  f(x) =(sin((pix)/2))/(x^3-2x^2+x) ?

Jun 1, 2016

$f \left(x\right) = \sin \frac{\frac{\pi x}{2}}{{x}^{3} - 2 {x}^{2} + x}$ has a hole at $x = 0$ and vertical asymptote at $x = 1$.

#### Explanation:

f(x)=sin((pix)/2)/(x^3-2x^2+x)=sin((pix)/2)/(x(x^2-2x+1)

= $\sin \frac{\frac{\pi x}{2}}{x {\left(x - 1\right)}^{2}}$

Hence $L {t}_{x \to 0} f \left(x\right) = L {t}_{x \to 0} \sin \frac{\frac{\pi x}{2}}{x {\left(x - 1\right)}^{2}}$

= $\frac{\pi}{2} L {t}_{x \to 0} \sin \frac{\frac{\pi x}{2}}{\left(\frac{\pi x}{2}\right) {\left(x - 1\right)}^{2}}$

= $L {t}_{x \to 0} \sin \frac{\frac{\pi x}{2}}{\frac{\pi x}{2}} \times L {t}_{x \to 0} \frac{1}{x - 1} ^ 2 = \frac{\pi}{2} \times 1 \times 1 = \frac{\pi}{2}$

It is apparent that at $x = 0$, the function is not defined, though it has a value of $\frac{\pi}{2}$, hence it has a hole at $x = 0$

Further it has vertical asymptote at $x - 1 = 0$ or $x = 1$

graph{sin((pix)/2)/(x(x-1)^2) [-8.75, 11.25, -2.44, 7.56]}