Question

find the following limit or state that it does not exist.

\\\lim_{x \to -1} \frac{3(2x-2)^2-48}{x+1}\\

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

a. \\\lim_{x \to -1} \frac{3(2x-2)^2-48}{x+1} = \\
(type an exact answer.)

b. the limit does not exist.

Explanation:

Evaluate direct substitution

$$ \lim_{x \to -1} \frac{3(2x-2)^2 - 48}{x+1} $$

Substituting \(x = -1\):

$$ \text{Numerator: } 3(2(-1)-2)^2 - 48 = 3(-4)^2 - 48 = 3(16) - 48 = 0 $$

$$ \text{Denominator: } -1 + 1 = 0 $$

This yields the indeterminate form \(\frac{0}{0}\).

Simplify the algebraic expression

Factor the numerator to cancel the common factor \(x+1\):

$$
LATEXBLOCK0
$$

Compute the limit

Substitute the factored form back into the limit:

$$
LATEXBLOCK1
$$

Answer:

• A. \(\lim_{x \to -1} \frac{3(2x-2)^2-48}{x+1} = -48\) (Correct answer)
• B. The limit does not exist.