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

1 Answer
Jun 1, 2016

Answer:

#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]}