Question

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

Answer 1

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

Hence `Lt_(x->0)f(x)=Lt_(x->0)sin((pix)/2)/(x(x-1)^2)`

= `pi/2Lt_(x->0)sin((pix)/2)/(((pix)/2)(x-1)^2)`

= `Lt_(x->0)sin((pix)/2)/((pix)/2)xxLt_(x->0)1/(x-1)^2=pi/2xx1xx1=pi/2`

It is apparent that at `x=0`, the function is not defined, though it has a value of `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]}