#### Answer

$$T_{2}(x) =(x-1)-\frac{3}{2}(x-1)^{2}$$
$$T_{3}(x) =(x-1)-\frac{3}{2}(x-1)^{2}+\frac{11}{6}(x-1)^{3}$$

#### Work Step by Step

Given $$f(x)=\frac{\ln x}{x}, \quad a=1$$
Since
\begin{aligned}
f(x) &=\frac{\ln x}{x} & & f(a)=0 \\
f^{\prime}(x) &=\frac{1-\ln x}{x^{2}} & & f(a)=1 \\
f^{\prime \prime}(x) &=\frac{-3+2 \ln x}{x^{3}} & f(a) &=-3 \\
f^{\prime \prime \prime}(x) &=\frac{11-6 \ln x}{x^{4}} & f(a) &=11
\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-1)+\frac{-3}{2 !}(x-1)^{2} \\
&=(x-1)-\frac{3}{2}(x-1)^{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-1)+\frac{-3}{2 !}(x-1)^{2}+\frac{11}{3 !}(x-1)^{3} \\
&=(x-1)-\frac{3}{2}(x-1)^{2}+\frac{11}{6}(x-1)^{3}
\end{aligned}