Answer
$A_{0}(x)=0 \\ A_{1}(x)=x-1 \\A_{2}(x)=x-1-\dfrac{(x-1)^2}{2} \\A_{3}(x)=x-1-(\dfrac{1}{2})(x-1)^2+(\dfrac{1}{3})(x-1)^3$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(1)=0 , \\f'(x)=\dfrac{1}{x} \implies f'(1)=1 \\ f''(x)=-\dfrac{1}{x^2} \implies f''(1)=-1 \\ f'''(x)=\dfrac{2}{x^3} \implies f'''(1)=2$
This implies that $A_{0}(x)=0 \\ A_{1}(x)=x-1 \\A_{2}(x)=x-1-\dfrac{(x-1)^2}{2} \\A_{3}(x)=x-1-(\dfrac{1}{2})(x-1)^2+(\dfrac{1}{3})(x-1)^3$
