use the limit definition of the derivative to find the following: if $f(x)=\frac{4}{x^{2}}$, find $f(4)$.

Question

Accepted Answer

# Explanation:
## Step1: Recall limit - definition of derivative
The limit - definition of the derivative is $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 4$ and $f(x)=\frac{4}{x^{2}}$, so $f(4)=\frac{4}{4^{2}}=\frac{4}{16}=\frac{1}{4}$, and $f(4 + h)=\frac{4}{(4 + h)^{2}}$.
## Step2: Substitute into the limit - definition
$$
\begin{align*}
f^{\prime}(4)&=\lim_{h
ightarrow0}\frac{\frac{4}{(4 + h)^{2}}-\frac{1}{4}}{h}\
&=\lim_{h
ightarrow0}\frac{\frac{16-(4 + h)^{2}}{4(4 + h)^{2}}}{h}\
&=\lim_{h
ightarrow0}\frac{16-(16 + 8h+h^{2})}{4h(4 + h)^{2}}\
&=\lim_{h
ightarrow0}\frac{16 - 16-8h - h^{2}}{4h(4 + h)^{2}}\
&=\lim_{h
ightarrow0}\frac{-8h - h^{2}}{4h(4 + h)^{2}}
\end{align*}
$$
## Step3: Simplify the expression
Cancel out the common factor $h$ in the numerator and denominator:
$$
\begin{align*}
f^{\prime}(4)&=\lim_{h
ightarrow0}\frac{-8 - h}{4(4 + h)^{2}}
\end{align*}
$$
## Step4: Evaluate the limit
Substitute $h = 0$ into the simplified expression:
$$
\begin{align*}
f^{\prime}(4)&=\frac{-8-0}{4(4 + 0)^{2}}\
&=\frac{-8}{4\times16}\
&=-\frac{1}{8}
\end{align*}
$$
# Answer:
$-\frac{1}{8}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall limit - definition of derivative

Step2: Substitute into the limit - definition

Step3: Simplify the expression

Step4: Evaluate the limit

Answer: