Examine the following function for continuity: $f(x) = \frac{1}{x-5}, x \neq 5$.
(N/A) The given function is $f(x) = \frac{1}{x-5}$,where $x \neq 5$.
For any real number $k$ such that $k \neq 5$,we evaluate the limit:
$\lim_{x \to k} f(x) = \lim_{x \to k} \frac{1}{x-5} = \frac{1}{k-5}$.
Also,the value of the function at $x = k$ is $f(k) = \frac{1}{k-5}$.
Since $\lim_{x \to k} f(x) = f(k)$ for all $k \in \mathbb{R} \setminus \{5\}$,the function $f$ is continuous at every point in its domain.
Therefore,$f$ is a continuous function.
For any real number $k$ such that $k \neq 5$,we evaluate the limit:
$\lim_{x \to k} f(x) = \lim_{x \to k} \frac{1}{x-5} = \frac{1}{k-5}$.
Also,the value of the function at $x = k$ is $f(k) = \frac{1}{k-5}$.
Since $\lim_{x \to k} f(x) = f(k)$ for all $k \in \mathbb{R} \setminus \{5\}$,the function $f$ is continuous at every point in its domain.
Therefore,$f$ is a continuous function.
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