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