Examine the following function for continuity: $f(x) = \frac{x^{2} - 25}{x + 5}, x \neq -5$.
(N/A) The given function is $f(x) = \frac{x^{2} - 25}{x + 5}$ for $x \neq -5$.
For any real number $c \neq -5$,we have:
$\lim_{x \to c} f(x) = \lim_{x \to c} \frac{x^{2} - 25}{x + 5} = \lim_{x \to c} \frac{(x + 5)(x - 5)}{x + 5} = \lim_{x \to c} (x - 5) = c - 5$.
Also,$f(c) = \frac{c^{2} - 25}{c + 5} = \frac{(c + 5)(c - 5)}{c + 5} = c - 5$.
Since $\lim_{x \to c} f(x) = f(c)$ for all $c \neq -5$,the function $f$ is continuous at every point in its domain.
For any real number $c \neq -5$,we have:
$\lim_{x \to c} f(x) = \lim_{x \to c} \frac{x^{2} - 25}{x + 5} = \lim_{x \to c} \frac{(x + 5)(x - 5)}{x + 5} = \lim_{x \to c} (x - 5) = c - 5$.
Also,$f(c) = \frac{c^{2} - 25}{c + 5} = \frac{(c + 5)(c - 5)}{c + 5} = c - 5$.
Since $\lim_{x \to c} f(x) = f(c)$ for all $c \neq -5$,the function $f$ is continuous at every point in its domain.
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