# How do you find the limit of (x-5)/(x^2-25) as x->5^-?

Nov 9, 2016

${\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25} = \frac{1}{10}$

#### Explanation:

${\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25} = {\lim}_{x \to {5}^{-}} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}} \left(x + 5\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25}} = {\lim}_{x \to {5}^{-}} \frac{1}{x + 5}$

$\textcolor{w h i t e}{{\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25}} = \frac{1}{5 + 5}$

$\textcolor{w h i t e}{{\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25}} = \frac{1}{10}$

Nov 9, 2016

$\frac{1}{10}$

#### Explanation:

Substituting $5$ in the given expression results to an indeterminate solution:

${\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25}$

$= \frac{5 - 5}{{5}^{2} - 25}$

$= \frac{0}{0} \text{ }$ Indeterminate solution.

let us think about another way to find the limit.

Here, Factorizing ${x}^{2} - 25$ is the best way.

Factorization ${x}^{2} - 25 \text{ }$is computed by applying the

polynomial identity.

$\textcolor{b l u e}{\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)}$

$\textcolor{b l u e}{{x}^{2} - 25 = {x}^{2} - {5}^{2} = \left(x - 5\right) \left(x + 5\right)}$

${\lim}_{x \to {5}^{-}} \frac{x - 5}{{x}^{2} - 25}$

$= {\lim}_{x \to {5}^{-}} \frac{\cancel{x - 5}}{\textcolor{b l u e}{\cancel{x - 5} \left(x + 5\right)}}$

$= {\lim}_{x \to {5}^{-}} \frac{1}{x + 5}$

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

$= \frac{1}{10}$