Question
Determine the value of ‘k’ for which the following function is continuous at x = 3:
$\text{f(x)} = \begin{cases} \frac{(\text{x}+3)^2-36}{\text{x}-3}\ \ \ \ ,\ \text{x}\neq3\\ \ \ \ \ \ \ \text{k}\ \ \ \ \ \ \ \ \ \ \ ,\ \text{x}=3 \end{cases}$
$\text{f(x)} = \begin{cases} \frac{(\text{x}+3)^2-36}{\text{x}-3}\ \ \ \ ,\ \text{x}\neq3\\ \ \ \ \ \ \ \text{k}\ \ \ \ \ \ \ \ \ \ \ ,\ \text{x}=3 \end{cases}$