Question

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

Answer 1

The vertical asymptotes of `y=secx` are

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

which look like this (in red).

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.