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What are the asymptotes for # g(x)= (x^3+1)/(x+1)#?

Answer 1
Christian AntunezChristian Antunez

#g(x) = (x^3+1)/(x+1) = ((x+1)(x^2-x+1))/(x+1) = x^2-x+1#

with exclusion #x != -1#

The polynomial #x^2-x+1# has no asymptotes, so neither has #g(x)#

The sum of cubes identity is:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

So

#x^3+1 = x^3+1^3 = (x+1)(x^2-x+1)#

So

#g(x) = (x^3+1)/(x+1) = ((x+1)(x^2-x+1))/(x+1) = x^2-x+1#
with exclusion #x != -1#
The polynomial #x^2-x+1# has no linear asymptotes, so neither has #g(x)#.
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Answer 2
Elijah AllenElijah Allen

The function g(x) = (x^3 + 1)/(x + 1) has a vertical asymptote at x = -1 and no horizontal asymptote.

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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.

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.

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.

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.

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