Determine the value of '$k$' for which the following function is continuous at $x = 3$: $f(x) = \begin{cases} \frac{(x + 3)^2 - 36}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases}$

$k = 12$

The correct answer is Option (3) → $k = 12$ ##

$f(x) = \begin{cases} \frac{(x + 3)^2 - 36}{x - 3} & ; x \neq 3 \\ k & ; x = 3 \end{cases}$

$f(x)$ is continuous at $x = 3$

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^+} f(x) = f(3)$

$\lim\limits_{x \to 3} \frac{(x + 3)^2 - (6)^2}{x - 3} = k$

$\lim\limits_{x \to 3} \frac{(x + 3 - 6)(x + 3 + 6)}{(x - 3)} = k$

$\lim\limits_{x \to 3} \frac{(x - 3)(x + 9)}{(x - 3)} = k$

$\lim\limits_{x \to 3} (x + 9) = k$

$12 = k$

$k = 12$