Question

appropriate. show your
points.
\\( \lim _ { h \
ightarrow 0 } \frac { 3 ( 2 + h ) ^ { 2 } - 12 } { h } \\)
3(4 + 4h + h ^ { 2 } ) - 12
\\( \frac { ( 12 + 12 h + 3 h ^ { 2 } ) - 12 } { } \\)

Explanation:

Step1: Expand the numerator

$$\begin{align*} 3(2 + h)^2-12&=3(4 + 4h+h^2)-12\\ &=12 + 12h+3h^2-12\\ &=12h + 3h^2 \end{align*}$$

Step2: Simplify the fraction

The original limit $\lim_{h
ightarrow0}\frac{3(2 + h)^2-12}{h}$ becomes $\lim_{h
ightarrow0}\frac{12h + 3h^2}{h}$. Factor out an $h$ from the numerator: $\lim_{h
ightarrow0}\frac{h(12 + 3h)}{h}$. Since $h
eq0$ as $h
ightarrow0$ (we are looking at the limit, not the value at $h = 0$), we can cancel the $h$ terms. So we get $\lim_{h
ightarrow0}(12 + 3h)$.

Step3: Evaluate the limit

Substitute $h = 0$ into $12+3h$. We have $12+3\times0=12$.

Answer:

$12$