How do you find a vertical asymptote for y = sec(x)?

1 Answer
Sep 25, 2014

The vertical asymptotes of #y=secx# are

#x={(2n+1)pi}/2#, where #n# is any integer,

which look like this (in red).

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Let us look at some details.

#y=secx=1/{cosx}#

In order to have a vertical asymptote, the (one-sided) limit has to go to either #infty# or #-infty#, which happens when the denominator becomes zero there.

So, by solving

#cosx=0#

#Rightarrow x=pm pi/2, pm{3pi}/2, pm{5pi}/2, ...#

#Rightarrow x=pi/2+npi={(2n+1)pi}/2#, where #n# is any integer.

Hence, the vertical asymptotes are

#x={(2n+1)pi}/2#, where #n# is any integer.