AP Calculus AB and BC

Limit of (x^3 + 1)/(x + 1) at -1 Is 3

The limit of x cubed plus 1, over x plus 1, as x approaches negative 1 is 3. The sum of cubes splits the numerator into x plus 1 times x squared minus x plus 1. Cancelling x plus 1 leaves the quadratic, and substituting negative 1 gives 1 plus 1 plus 1.

lim⁡x→−1x3+1x+1=3\lim_{x \to -1} \frac{x^{3}+1}{x+1} = 3

Settled by factoring the sum of cubes.

The sum of cubes does the work

a3+b3=(a+b)(a2−ab+b2)a^{3}+b^{3} = (a+b)\left(a^{2}-ab+b^{2}\right)

Taking a=xa = x and b=1b = 1 turns the numerator into (x+1)(x2−x+1)(x+1)\left(x^{2}-x+1\right), which shares the factor x+1x+1 with the denominator.

lim⁡x→−1(x+1)(x2−x+1)x+1=lim⁡x→−1(x2−x+1)=1+1+1=3\lim_{x \to -1}\frac{(x+1)\left(x^{2}-x+1\right)}{x+1} = \lim_{x \to -1}\left(x^{2}-x+1\right) = 1+1+1 = 3

Getting the middle sign right

Most students meet the difference of cubes first, a3−b3=(a−b)(a2+ab+b2)a^{3}-b^{3} = (a-b)\left(a^{2}+ab+b^{2}\right), and then stall on a sum. It is the same formula with bb replaced by −b-b, which flips two signs and leaves the outer terms alone.

The pattern in words

The bracket with two terms carries the sign of the original. The bracket with three terms takes the opposite sign in its middle term, and its outer terms are always positive. For a sum of cubes that reads plus, then minus, then plus.

Check by expanding: (x+1)(x2−x+1)=x3−x2+x+x2−x+1=x3+1(x+1)\left(x^{2}-x+1\right) = x^{3}-x^{2}+x+x^{2}-x+1 = x^{3}+1. If the middle sign is wrong, the cross terms fail to cancel and the check catches it in one line.

The mistakes students make

The first two are sign or structure slips that land on a definite wrong number. The third is a misreading of the indeterminate form.

  • Using x2+x+1x^{2}+x+1 as the second factor and answering 11. That bracket belongs to x3−1x^{3}-1.
  • Treating x3+1x^{3}+1 as (x+1)3(x+1)^{3}, cancelling to (x+1)2(x+1)^{2} and answering 00.
  • Reporting no limit on the strength of 00\frac{0}{0}. That form says substitution failed, not that the limit is missing.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of x3+1x+1\frac{x^{3}+1}{x+1} at x = -1?

It is 33.

How do you factor x cubed plus 1?

x3+1=(x+1)(x2−x+1)x^{3}+1 = (x+1)\left(x^{2}-x+1\right). The signs inside the second bracket run plus, minus, plus.

Does L'Hopital's rule give the same answer?

Yes. Differentiating top and bottom gives 3x21\frac{3x^{2}}{1}, which is 33 at x=−1x = -1. Factoring is the Unit 1 method and needs no derivatives.