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

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

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Answer 1 · 2014-09-25T06:50:38

The vertical asymptotes of $y=secx$ are

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

which look like this (in red).

enter image source here

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.

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

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