## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 5}\dfrac{x^{2}-6x+5}{x-5}=4$
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: $\lim\limits_{x \to 5}\dfrac{x^{2}-6x+5}{x-5}=\lim\limits_{x \to 5}\dfrac{(x-5)(x-1)}{x-5}=\lim\limits_{x \to 5}x-1=5-1=4$