1952 AMC 12 Problem 36

Question

To be continuous at

x=-1,

the value of

\dfrac{x^3+1}{x^2-1}

is taken to be:

-2

0

\dfrac32

\infty

-\dfrac32

Factor both numerator and denominator before substituting

x=-1

Cancel the common factor

x+1

and evaluate the remaining expression

For

x\ne-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=-1

is

\frac{1+1+1}{-2}=-\frac32.

Assigning this value removes the discontinuity.

Thus, the correct answer is E.

Problem 36 in other years