Answer
$$T_3(x) =\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2)^{2}-\frac{1}{16}(x-2)^{3} $$
Work Step by Step
Given $$f(x) =\frac{1}{x},\ \ a=2$$
Since
\begin{align*}
f(x) &=\frac{1}{x}\ \ \ \ \ \ \ \ \ \ f(2) =\frac{1}{2}\\
f'(x) &=\frac{-1}{x^2}\ \ \ \ \ \ \ f'(2 )=\frac{-1}{4}\\
f''(x) &=\frac{2}{x^3}\ \ \ \ \ \ \ f''(2 )=\frac{1}{4}\\
f'''(x) &=\frac{-6}{x^4}\ \ \ \ \ \ \ f''(2 )=\frac{-3}{8}
\end{align*}
Then
\begin{align*}
T_3(x)&= \sum_{n=0}^{3}\frac{f^{(n)}(2)}{n!}(x-2)^n\\
& =\frac{\frac{1}{2}}{0 !}-\frac{1}{1 !}(x-2)+\frac{1}{2 !}(x-2)^{2}-\frac{\frac{3}{3}}{3 !}(x-2)^{3} \\
&=\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2)^{2}-\frac{1}{16}(x-2)^{3}
\end{align*}
