Answer
$8$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 4} \dfrac{x^2-16}{\ln (x-3)}$
Recall the series for $\ln (1+x)=\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{(x)^{k}}{k}$ and $\ln (x-3)=\ln (x-4+1)=\ln [1+(x-4)]=\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{(x-4)^{k}}{k}$
Now, $L=\lim\limits_{x \to 4} \dfrac{(x-4)(x+4)}{\ln (x-3)}\\=\lim\limits_{x \to 4} \dfrac{(x-4)(x+4)}{\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{(x-4)^{k}}{k}}\\=\lim\limits_{x \to 4} \dfrac{(x-4)(x+4)}{ (-1)^{1+1} \dfrac{(x-4)^{1-1}}{1}+(-1)^{2+1} \dfrac{(x-4)^{2-1}}{1}+.......}\\=\lim\limits_{x \to 4} \dfrac{x+4}{\dfrac{1}{1}-\dfrac{x-4}{2}+\dfrac{(x-4)^2}{3}+.........}\\=\dfrac{4+4}{1-\dfrac{4-4}{2}+\dfrac{(4-4)^2}{2}+........}\\=\dfrac{8}{1-0+0-0+.....}\\=8$
