Question

The limit of $\dfrac{x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:

$0$

Indeterminate

$x-1$

$2$

$1$

Answer: D

Concepts:calculus difference of squares

Difficulty rating: 1470

Small Hint:

Factor the numerator as a difference of squares

Big Hint:

For $x\ne1,$ cancel the common factor $x-1$

Solution:

For $x\ne1,$ $\begin{aligned} \frac{x^2-1}{x-1} &=\frac{(x-1)(x+1)}{x-1}\\ &=x+1. \end{aligned}$ Therefore the limit as $x$ approaches $1$ is $1+1=2.$

Thus, the correct answer is D.

1950 AMC 12 Problem 40

Answer: D

Difficulty rating: 1470

Small Hint:

Big Hint:

Solution:

Problem 40 in Other Years