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

Answer 1

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

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#.

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Answer 2

Avery Alexander

The function f(x) = secx has vertical asymptotes at x = π/2 + nπ and x = -π/2 + nπ, where n is an integer. There are no horizontal asymptotes or holes in the graph of f(x) = secx.

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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.