#### Answer

$\dfrac{1}{10}$

#### Work Step by Step

We want to find $\lim\limits_{x \to 5}\dfrac{x-5}{x^2-25}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 5 is zero, and if we try to substitute $x=5$ directly into our function, we get zero in both the numerator and the denominator. This means we have a common factor of $x-5$ in both the numerator and the denominator.
Note that $\dfrac{x-5}{x^2-25}=\dfrac{x-5}{(x-5)(x+5)}=\dfrac{1}{x+5}$ for all $x \neq 5.$
Thus $\lim\limits_{x \to 5}\dfrac{x-5}{x^2-25}=\lim\limits_{x \to 5}\dfrac{1}{x+5}=\dfrac{1}{5+5}=\dfrac{1}{10}.$