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What are the asymptote(s) and hole(s), if any, of # f(x) =1/(x^2+2) #?

Answer 1

#f(x)# has a horizontal asymptote #y=0# and no holes

#x^2 >= 0# for all #x in RR#

So #x^2+2 >= 2 > 0# for all #x in RR#

That is, the denominator is never zero and #f(x)# is well defined for all #x in RR#, but as #x->+-oo#, #f(x) -> 0#. Hence #f(x)# has a horizontal asymptote #y=0#.

graph{1/(x^2+2) [-2.5, 2.5, -1.25, 1.25]}

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Answer 2

Nathan Axel

The function f(x) = 1/(x^2+2) has no holes and two vertical asymptotes at x = √2 and x = -√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.