For what values of x, if any, does #f(x) = 1/((x^2-4)cos(pi/2+(8pi)/x) # have vertical asymptotes?

1 Answer
Mar 31, 2017

Answer:

  • There is an essential singularity when #x=0#
  • There will be vertical asymptotes at #x = +-2 #
  • There will be an infinite number of vertical asymptotes at # x =8/n \ \ n in ZZ//0 #. (E.g. # x=+-8, x=+-4, x=+-8/3, +-2, +-8/5, +-4/3, +-8/7, +-1 , ... #)

Explanation:

We have:

# f(x) = 1/((x^2-4)cos(pi/2+(8pi)/x) #

There will be vertical asymptotes if

# (x^2-4) = 0 => x = +-2 #

There will also be vertical asymptotes if:

# cos(pi/2+(8pi)/x) =0 #
# :. pi/2+(8pi)/x =+-pi/2, +-(3pi)/2, ... #
# :. pi/2+(8pi)/x =(2n+1)pi/2 \ \ n in ZZ#
# :. pi/2+(8pi)/x =npi + pi/2#
# :. (8pi)/x =npi#
# :. 8/x =n#
# :. x =8/n \ \ n ne 0 #

With:

# n= +- 1 => x=+-8 #
# n= +- 2 => x=+-4 #
# n= +- 3 => x=+-8/3 #
# n= +- 4 => x=+-2 #
# n= +- 5 => x=+-8/5 #
# n= +- 6 => x=+-4/3 #
# n= +- 7 => x=+-8/7 #
# n= +- 8 => x=+-1 #
# vdots #

So we have an infinite number of asymptotes. As #n rarr oo# then #8/n rarr 0# so as we approach the origin we have an increasing number of closely spaced asymptotes, until in the limit we have an infinite number of asymptotes at the singularity itself.

We also need to consider the argument to the cosine function itself:

# pi/2+(8pi)/x #

Which is invalid at #x=0#, There is an essential singularity when #x=0#

This is a graph of the function which shows that as #n# increases the asymptotes occur at #8/n#

graph{1/((x^2-4)cos(pi/2+(8pi)/x) [-12, 12, -1.25, 1.25]}