Question

use the properties of limits to help decide whether the limit exists. if the limit exists, find its value. $lim_{x
ightarrow1}\frac{x^{2}-1}{x - 1}$ select the correct choice below and, if necessary, fill in the answer box within your choice. a. $lim_{x
ightarrow1}\frac{x^{2}-1}{x - 1}=square$ (simplify your answer.) b. the limit does not exist and is neither $infty$ nor $-infty$.

Explanation:

Step1: Factor the numerator

We know that $x^{2}-1=(x + 1)(x - 1)$. So the limit $\lim_{x
ightarrow1}\frac{x^{2}-1}{x - 1}=\lim_{x
ightarrow1}\frac{(x + 1)(x - 1)}{x - 1}$.

Step2: Cancel out the common factor

Since $x
eq1$ when taking the limit (we are approaching 1, not equal to 1), we can cancel out the $(x - 1)$ terms. So $\lim_{x
ightarrow1}\frac{(x + 1)(x - 1)}{x - 1}=\lim_{x
ightarrow1}(x + 1)$.

Step3: Evaluate the limit

Substitute $x = 1$ into $x+1$. We get $\lim_{x
ightarrow1}(x + 1)=1 + 1=2$.

Answer:

A. $\lim_{x
ightarrow1}\frac{x^{2}-1}{x - 1}=2$