Q.Find the value of $k$ for which $f(x) = \dfrac{x^2 - 25}{x - 5}$ (for $x \neq 5$), $f(5) = k$, is continuous at $x = 5$.

Continuity at $x=5$ requires $k=f(5)$ to equal the limit of $f$ as $x\to5$.

Continuity at $x = 5$ requires $k = f(5) = \displaystyle\lim_{x\to 5} f(x)$ (§6, Case A).

Compute the limit. For $x \neq 5$, $\dfrac{x^2 - 25}{x - 5} = \dfrac{(x-5)(x+5)}{x-5} = x + 5$. Hence $\displaystyle\lim_{x\to 5} f(x) = 5 + 5 = 10$.

Set the constant. $k = 10$. …

Substituting $x=5$ directly into $\dfrac{x^2-25}{x-5}$ gives the indeterminate $\tfrac00$; the factor $x-5$ must be cancelled before evaluating. Settin …