Answer
$\dfrac{1}{8}$
Work Step by Step
Write the Taylor series for the function as: $\ln x=x-\dfrac{(x)^2}{2}+\dfrac{(x)^3}{3}+......$
Now, we have:
$\lim\limits_{x \to 4} \dfrac{\ln (x-4)}{x^2-16}=\lim\limits_{x \to 4} \dfrac{(x-4)-\dfrac{(x-4)^2}{2}+\dfrac{(x-4)^3}{3!}+......}{x^2-16}
\\=\lim\limits_{x \to 4} \dfrac{(x-4)-\dfrac{(x-4)^2}{2}+\dfrac{(x-4)^3}{3!}+......}{(x-4)(x+4)}\\=\lim\limits_{x \to 4} \dfrac{1-\dfrac{(x-4)}{2}+\dfrac{(x-4)^2}{3!}+......}{x+4}\\=\dfrac{1}{8}$
