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

1 Answer
Dec 8, 2017

Answer:

#f(x)# 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(x) = cos((pix)/2)/((x-1)(x+2))#

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

As #x->1#, then #cos((pix)/2) -> 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->1) f(x) = lim_(t->0) cos((pit)/2+pi/2)/(t(t+3))#

#color(white)(lim_(x->1) f(x)) = lim_(t->0) -sin((pit)/2)/((pit)/2) * ((pi/2)/(t+3))#

#color(white)(lim_(x->1) f(x)) = -1 * pi/6#

#color(white)(lim_(x->1) f(x)) = -pi/6#

So #f(x)# has a hole at #x=1#

As #x->-2#, then #cos((pix)/2)->cos(-pi) = -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 #abs(cos((pix)/2)) <= 1#, while #(x-1)(x+2) -> oo# as #x -> +-oo#.

Hence #f(x)# has a horizontal asymptote #y=0#.

Note that historically some people would not count this as an asymptote since the graph of #f(x)# 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]}