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

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(a)=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}$. 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\rightarrow0}\frac{\frac{4}{(4 + h)^{2}}-\frac{1}{4}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{16-(4 + h)^{2}}{4(4 + h)^{2}}}{h}\ &=\lim_{h\rightarrow0}\frac{16-(16 + 8h+h^{2})}{4h(4 + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{16 - 16-8h - h^{2}}{4h(4 + h)^{2}}\ &=\lim_{h\rightarrow0}\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\rightarrow0}\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:

$-\frac{1}{8}$