Question

given the function $f(x)=\frac{1}{3x}$, determine the instantaneous rate of change of $f$ at $x = 5$ using the limit shown below. you do not have to simplify your answer.

Explanation:

Step1: Recall the formula for instantaneous rate of change

The instantaneous rate of change of a function $y = f(x)$ at $x=a$ is given by $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 5$ and $f(x)=\frac{1}{3x}$. So, $f(5)=\frac{1}{3\times5}=\frac{1}{15}$ and $f(5 + h)=\frac{1}{3(5 + h)}$.

Step2: Substitute into the limit formula

\[

$$\begin{align*} f^{\prime}(5)&=\lim_{h ightarrow0}\frac{f(5 + h)-f(5)}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{1}{3(5 + h)}-\frac{1}{15}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{15-(5 + h)}{45(5 + h)}}{h}\\ &=\lim_{h ightarrow0}\frac{15-(5 + h)}{45h(5 + h)}\\ &=\lim_{h ightarrow0}\frac{15 - 5 - h}{45h(5 + h)}\\ &=\lim_{h ightarrow0}\frac{10 - h}{45h(5 + h)} \end{align*}$$

\]

Answer:

$\lim_{h
ightarrow0}\frac{\frac{1}{3(5 + h)}-\frac{1}{15}}{h}$ or $\lim_{h
ightarrow0}\frac{10 - h}{45h(5 + h)}$