Question

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

a. (limlimits_{x \to 4} \frac{4x^2 - 64}{x - 4} = square) (type an integer or a simplified fraction.)

b. the limit does not exist and is neither (infty) nor (-infty).

Explanation:

Step1: Factor the numerator

First, factor the numerator \(4x^2 - 64\). We can factor out a 4 first: \(4(x^2 - 16)\). Then, \(x^2 - 16\) is a difference of squares, so it factors to \((x - 4)(x + 4)\). So the numerator becomes \(4(x - 4)(x + 4)\).
The expression is now \(\lim_{x \to 4} \frac{4(x - 4)(x + 4)}{x - 4}\).

Step2: Cancel common factors

We can cancel the common factor of \(x - 4\) (since \(x \to 4\) but \(x
eq 4\) when taking the limit, so we can cancel this factor). After canceling, we have \(\lim_{x \to 4} 4(x + 4)\).

Step3: Substitute \(x = 4\)

Now, substitute \(x = 4\) into the expression \(4(x + 4)\). So we get \(4(4 + 4) = 4\times8 = 32\).

Answer:

A. \(\lim\limits_{x\to 4} \frac{4x^2 - 64}{x - 4} = \boxed{32}\)