Question

evaluate the limit: \\(\lim\limits_{h \to 0} \frac{(-4 + h)^2 - 16}{h}\\)

Explanation:

Step1: Expand the squared term

$$(-4+h)^2 = (-4)^2 + 2(-4)(h) + h^2 = 16 - 8h + h^2$$

Step2: Substitute into the limit

$$\lim_{h \to 0} \frac{16 - 8h + h^2 - 16}{h} = \lim_{h \to 0} \frac{-8h + h^2}{h}$$

Step3: Simplify the rational expression

$$\lim_{h \to 0} \frac{h(-8 + h)}{h} = \lim_{h \to 0} (-8 + h)$$

Step4: Evaluate the limit

Substitute $h=0$ into $-8+h$

Answer:

$-8$