Question

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

\\\lim_{x \to -1} \frac{4x^2 + 3x - 1}{4x - 1}\\

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

a. \\\lim_{x \to -1} \frac{4x^2 + 3x - 1}{4x - 1} = \quad\\ (simplify your answer.)
b. the limit does not exist.

Explanation:

Identify the function and limit point

$$ f(x) = \frac{4x^2 + 3x - 1}{4x - 1}, \quad c = -1 $$

Evaluate the denominator at the limit point

$$ 4(-1) - 1 = -5 eq 0 $$

Apply direct substitution

$$ \lim_{x \to -1} \frac{4x^2 + 3x - 1}{4x - 1} = \frac{4(-1)^2 + 3(-1) - 1}{4(-1) - 1} = \frac{4 - 3 - 1}{-5} = \frac{0}{-5} = 0 $$

Answer:

• (A) \(\lim_{x \to -1} \frac{4x^2 + 3x - 1}{4x - 1} = 0\) (Correct answer)
• (B) The limit does not exist.