What are the asymptote(s) and hole(s), if any, of # f(x) = secx#?

1 Answer
Jan 2, 2018

Answer:

There are vertical asymptotes at #x=pi/2+pik,k in ZZ#

Explanation:

To look at this problem I will use the identity:
#sec(x)=1/cos(x)#

From this we see that there will be vertical asymptotes whenever #cos(x)=0#. Two values for when this occurs spring to mind, #x=pi/2# and #x=(3pi)/2#. Since the cosine function is periodic, these solutions will repeat every #2pi#.

Since #pi/2# and #(3pi)/2# only differ by #pi#, we can write all these solutions like this:
#x=pi/2+pik#, where #k# is any integer, #k in ZZ#.

The function has no holes, since holes would require both the numerator and the denominator to equal #0#, and the numerator is always #1#.