Question

determine the following limit, using \\(\infty\\) or \\(-\infty\\) when appropriate, or state that it does not exist. \\(\lim_{x \to 1} \frac{5x^2 - 5}{x - 1}\\) simplify the expression inside the limit, if possible. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. \\(\lim_{x \to 1} \frac{5x^2 - 5}{x - 1} = \lim_{x \to 1} \left( \square \
ight)\\) b. the expression inside the limit cannot be simplified.

Explanation:

Step1: Factor the numerator

First, factor the numerator \(5x^2 - 5\). We can factor out a 5: \(5x^2 - 5 = 5(x^2 - 1)\). Then, using the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\), we have \(x^2 - 1 = (x - 1)(x + 1)\). So, \(5x^2 - 5 = 5(x - 1)(x + 1)\).

Step2: Simplify the fraction

Now, substitute the factored form of the numerator into the fraction \(\frac{5x^2 - 5}{x - 1}\). We get \(\frac{5(x - 1)(x + 1)}{x - 1}\). Since \(x \to 1\) but \(x
eq 1\) (we are taking the limit as \(x\) approaches 1, not evaluating at \(x = 1\)), we can cancel out the common factor of \(x - 1\) in the numerator and the denominator. This leaves us with \(5(x + 1)\).

So, \(\lim_{x \to 1} \frac{5x^2 - 5}{x - 1} = \lim_{x \to 1} 5(x + 1)\)

Answer:

A. \(\lim\limits_{x\to 1}\frac{5x^2 - 5}{x - 1}=\lim\limits_{x\to 1} 5(x + 1)\) (the expression inside the limit simplifies to \(5(x + 1)\))