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1952 AMC 12 Problem 36

Attempt Problem 36 of the 1952 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1952 AMC 12 solutions, or check the answer key.

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36.

To be continuous at x=−1,x=-1, the value of x3+1x2−1\dfrac{x^3+1}{x^2-1} is taken to be:

−2-2

00

32\dfrac32

∞\infty

−32-\dfrac32

Answer: E
Concepts:calculusfactoring
Difficulty rating: 1450
Small Hint:

Factor both numerator and denominator before substituting x=−1x=-1

Big Hint:

Cancel the common factor x+1x+1 and evaluate the remaining expression

Solution:

For x≠−1,x\ne-1, x3+1x2−1=(x+1)(x2−x+1)(x+1)(x−1)=x2−x+1x−1. \begin{aligned} \frac{x^3+1}{x^2-1} &=\frac{(x+1)(x^2-x+1)} {(x+1)(x-1)}\\ &=\frac{x^2-x+1}{x-1}. \end{aligned} Its limit at x=−1x=-1 is 1+1+1−2=−32. \frac{1+1+1}{-2}=-\frac32. Assigning this value removes the discontinuity.

Thus, the correct answer is E.

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