Answer
$4$
Work Step by Step
The taylor series can be defined as: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$
We have
$\lim\limits_{x \to 2} \dfrac{x^2-4}{\ln (x-1)}=\lim\limits_{x \to 2} \dfrac{(x-2)(x+2)}{(x-2)-\dfrac{ (x-2)^2}{2}+\dfrac{(x-2)^3}{3}-....)} \\= \dfrac{ \lim\limits_{x \to 2} [(x-2)(x+2)] }{\lim\limits_{x \to 2} (x-2)-\dfrac{ (x-2)^2}{2}+\dfrac{(x-2)^3}{3}-....)} \\ =\dfrac{2+2}{1-0+0...}\\=4$
