To find the limit
[tex]\[
\lim _{x \rightarrow 5} \frac{x-5}{x^2-5x},
\][/tex]
we first notice that directly substituting [tex]\( x = 5 \)[/tex] into the expression yields an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. To resolve this, we must simplify the expression.
We start by factoring the denominator:
[tex]\[
x^2 - 5x = x(x - 5).
\][/tex]
Now, rewrite the original expression with this factorization:
[tex]\[
\frac{x-5}{x^2 - 5x} = \frac{x-5}{x(x-5)}.
\][/tex]
At this point, for [tex]\( x \neq 5 \)[/tex], we can cancel the common factor [tex]\( x - 5 \)[/tex] in the numerator and the denominator:
[tex]\[
\frac{x-5}{x(x-5)} = \frac{1}{x}.
\][/tex]
Now, we can straightforwardly identify the limit of the simplified expression as [tex]\( x \)[/tex] approaches 5:
[tex]\[
\lim_{x \to 5} \frac{1}{x} = \frac{1}{5}.
\][/tex]
Thus, the limit is:
[tex]\[
\boxed{\frac{1}{5}}.
\][/tex]
