use the properties of limits to help decide whether the limit exists. if the limit exists, find its value…

use the properties of limits to help decide whether the limit exists. if the limit exists, find its value. $lim_{x\rightarrow1}\frac{x^{2}-1}{x - 1}$ select the correct choice below and, if necessary, fill in the answer box within your choice. a. $lim_{x\rightarrow1}\frac{x^{2}-1}{x - 1}=square$ (simplify your answer.) b. the limit does not exist and is neither $infty$ nor $-infty$.

use the properties of limits to help decide whether the limit exists. if the limit exists, find its value. $lim_{x\rightarrow1}\frac{x^{2}-1}{x - 1}$ select the correct choice below and, if necessary, fill in the answer box within your choice. a. $lim_{x\rightarrow1}\frac{x^{2}-1}{x - 1}=square$ (simplify your answer.) b. the limit does not exist and is neither $infty$ nor $-infty$.

Answer

Explanation:

Step1: Factor the numerator

We know that $x^{2}-1=(x + 1)(x - 1)$. So the limit $\lim_{x\rightarrow1}\frac{x^{2}-1}{x - 1}=\lim_{x\rightarrow1}\frac{(x + 1)(x - 1)}{x - 1}$.

Step2: Cancel out the common factor

Since $x\neq1$ when taking the limit (we are approaching 1, not equal to 1), we can cancel out the $(x - 1)$ terms. So $\lim_{x\rightarrow1}\frac{(x + 1)(x - 1)}{x - 1}=\lim_{x\rightarrow1}(x + 1)$.

Step3: Evaluate the limit

Substitute $x = 1$ into $x+1$. We get $\lim_{x\rightarrow1}(x + 1)=1 + 1=2$.

Answer:

A. $\lim_{x\rightarrow1}\frac{x^{2}-1}{x - 1}=2$