Answer
$4$
Work Step by Step
First, apply direct substitution:
$\lim\limits_{x \to 5}\dfrac{x^{2}-6x+5}{x-5}=\dfrac{5^{2}-6(5)+5}{5-5}=\dfrac{0}{0}$ $(Indeterminate$ $Form)$
Apply factorization to the numerator:
$\lim\limits_{x \to 5}\dfrac{x^{2}-6x+5}{x-5}=\lim\limits_{x \to 5}\dfrac{(x-5)(x-1)}{x-5}=\lim\limits_{x \to 5}x-1=5-1=4$
