Question

find the following limit or state that it does not exist.
\\( \lim\limits_{x \to -1} \frac{4(2x - 1)^2 - 36}{x + 1} \\)

simplify the given limit
\\( \lim\limits_{x \to -1} \frac{4(2x - 1)^2 - 36}{x + 1} = \lim\limits_{x \to -1} \left( 16(x - 2) \
ight) \\) (simplify your answer)

evaluate the limit, if possible. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

\\( \boldsymbol{\circ} \\) a. \\( \lim\limits_{x \to -1} \frac{4(2x - 1)^2 - 36}{x + 1} = \square \\) (type an exact answer)
\\( \boldsymbol{\circ} \\) b. the limit does not exist.

Explanation:

Step1: Substitute \( x = -1 \) into \( 16(x - 2) \)

We know the simplified limit is \( \lim_{x \to -1} 16(x - 2) \). Now we substitute \( x = -1 \) into the expression \( 16(x - 2) \).
The expression becomes \( 16((-1)- 2) \)

Step2: Calculate the value

First, calculate the value inside the parentheses: \( (-1)-2=-3 \)
Then multiply by 16: \( 16\times(-3)= - 48 \)

Answer:

-48