How do you find the limit lim_(x->5)(x^2-6x+5)/(x^2-25) ?

Jul 29, 2014

$= \frac{2}{5}$

Explanation

$= {\lim}_{x \to 5} \frac{{x}^{2} - 6 x + 5}{{x}^{2} - 25}$ , if we plug $x = 5$, we get $\frac{0}{0}$ form

$= {\lim}_{x \to 5} \frac{{x}^{2} - 5 x - x + 5}{\left(x + 5\right) \left(x - 5\right)}$

$= {\lim}_{x \to 5} \frac{x \left(x - 5\right) - 1 \left(x - 5\right)}{\left(x + 5\right) \left(x - 5\right)}$

$= {\lim}_{x \to 5} \frac{\left(x - 5\right) \left(x - 1\right)}{\left(x + 5\right) \left(x - 5\right)}$

$= {\lim}_{x \to 5} \frac{\left(x - 1\right)}{\left(x + 5\right)}$

$= \frac{5 - 1}{5 + 5}$

$= \frac{4}{10}$$= \frac{2}{5}$