Answer
$\lim\limits_{x \to 5}\frac{(x-5)}{x^{2} - 25} = \frac{1}{10} = 0.1$
Work Step by Step
Simplify:
$\frac{(x-5)}{x^{2} - 25} = \frac{(x-5)}{(x-5)(x+5)} = \frac{1}{(x+5)}$
Now,
$\lim\limits_{x \to 5} \frac{(x-5)}{x^{2} - 25}= \lim\limits_{x \to 5}\frac{1}{(x+5)}$
= $\frac{1}{10}= 0.1$
