Answer
$$L(x)\approx -\frac{x}{10000}+\frac{1}{50}$$
$$0.01\%~error$$
Work Step by Step
Given $$\frac{1}{101} $$
Consider $f(x)=\dfrac{1}{x}, a=100,$ and $\Delta x=1$, since
\begin{align*}
f^{\prime}(x)&=-\frac{1}{x^2} \\
f^{\prime}(100)&=-\frac{1}{10000}
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=-\frac{1}{10000}(x-100)+\frac{1}{100}\\
&=-\frac{x}{10000}+\frac{1}{50}
\end{align*}
Since
\begin{align*}
L(101)&= f(101)\\
&=\frac{1}{101}\\
&\approx-\frac{101}{10000}+\frac{1}{50}\\
&\approx 0.0099
\end{align*}
Hence the error is given by
$$ | \frac{1}{101}-0.0099|=9.900\times 10^{-7}$$
and the percentage is
$$\frac{9.900\times 10^{-7} }{1/101}\times 100 \% \approx 0.01\%$$
