Q.Find the value of $k$ for which $f(x) = \dfrac{x^2 - 9}{x - 3}$ (for $x \neq 3$), $f(3) = k$, is continuous at $x = 3$.

For continuity at $x=3$, the value $k=f(3)$ must equal the limit of $f$ as $x\to3$.

Continuity at $x = 3$ requires $\displaystyle\lim_{x\to 3} f(x) = f(3) = k$ (§6, Case A). So compute the limit first.

Compute the limit. For $x \neq 3$, factor the numerator using $x^2 - 9 = (x-3)(x+3)$:

Hence $\displaystyle\lim_{x\to 3} f(x) = \lim_{x\to 3}(x + 3) = 3 + 3 = 6$.

Set the constant equal to the limit. For continuity, $k = f(3) = \displaystyle\lim_{x\to 3} f(x) = 6$.

So $f$ is continuous at $x = 3$ precisely when $k = 6$. …

A common mistake is to substitute $x=3$ directly into $\dfrac{x^2-9}{x-3}$, obtaining $\tfrac{0}{0}$, and either give up or wrongly set $k=0$. The indeterminate form must be resolved by cancell …