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What are the asymptote(s) and hole(s), if any, of # f(x) = 1/cosx#?

Answer 1

Hazel Anglin

There will be vertical asymptotes at #x = pi/2 + pin#, n and integer.

There will be asymptotes.

Whenever the denominator equals #0#, vertical asymptotes occur.

Let's set the denominator to #0# and solve.

#cosx = 0#

#x = pi/2, (3pi)/2#

Since the function #y = 1/cosx# is periodic, there will be infinite vertical asymptotes, all following the pattern #x = pi/2 + pin#, n an integer.

Finally, note that the function #y = 1/cosx# is equivalent to #y = secx#.

Hopefully this helps!

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Answer 2

Jace Anderson

The function f(x) = 1/cosx has a vertical asymptote at x = π/2 + nπ, where n is an integer. There are no horizontal asymptotes. There is a removable hole at x = π/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.