Question

1. use the limit definition of the derivative to find $g(-3)$ for $g(x)=\frac{1}{1 - x}$.

Explanation:

Step1: Recall the limit - definition of the derivative

The limit - definition of the derivative is $g^{\prime}(a)=\lim_{h
ightarrow0}\frac{g(a + h)-g(a)}{h}$. Here, $a=-3$ and $g(x)=\frac{1}{1 - x}$. So, $g(-3)=\frac{1}{1-(-3)}=\frac{1}{4}$, and $g(-3 + h)=\frac{1}{1-(-3 + h)}=\frac{1}{4 - h}$.

Step2: Substitute into the limit - definition

\[

$$\begin{align*} g^{\prime}(-3)&=\lim_{h ightarrow0}\frac{g(-3 + h)-g(-3)}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{1}{4 - h}-\frac{1}{4}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{4-(4 - h)}{4(4 - h)}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{4 - 4+h}{4(4 - h)}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{h}{4(4 - h)}}{h}\\ &=\lim_{h ightarrow0}\frac{h}{4h(4 - h)} \end{align*}$$

\]

Step3: Simplify the limit

Cancel out the $h$ terms: $\lim_{h
ightarrow0}\frac{h}{4h(4 - h)}=\lim_{h
ightarrow0}\frac{1}{4(4 - h)}$.

Step4: Evaluate the limit

Substitute $h = 0$ into $\frac{1}{4(4 - h)}$. We get $\frac{1}{4\times4}=\frac{1}{16}$.

Answer:

$\frac{1}{16}$