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What are the asymptote(s) and hole(s), if any, of # f(x) =1/x^2-1/(1-x)+x/(3-x)#?

Answer 1

Vertical asymptotes at #x={0,1,3}#

Asymptotes and holes are present due to the fact that the denominator of any fraction cannot be 0, since division by zero is impossible.

Since there are no cancelling factors, the non permissible values are all vertical asymptotes.

Therefore: #x^2=0# #x=0# and #3-x=0# #3=x# and #1-x=0# #1=x#

Which is all the vertical asymptotes.

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Answer 2

Isaiah Anthony

The function f(x) = 1/x^2 - 1/(1-x) + x/(3-x) has two vertical asymptotes at x = 0 and x = 1. There are no horizontal asymptotes. The function has a hole at x = 3.

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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.