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

##### 1 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

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

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

Which is invalid at

This is a graph of the function which shows that as

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