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