Answer
$6$
Work Step by Step
Since, we have $\lim_{x\to4}\dfrac{x^2-2x-8}{x-4}$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim_{x\to4}\dfrac{x^2-2x-8}{x-4}=\lim_{x\to4}\dfrac{\dfrac{d(x^2-2x-8)}{dx}}{\dfrac{d(x-4)}{dx}}$
or, $=\lim_{x\to4}\dfrac{2x-2}{1}$
or, $=\lim_{x\to4}(2x-2)$
or, $=2\times4-2$
or, $=6$
