​$$f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & \text{if } x \neq 3 \\5, & \text{if } x = 3\end{cases}$$, then f(x):

Correct option is B

Given:

​$$\begin{aligned}&\textbf{Simplify the Function for } x \neq 3 \\&\text{For } x \neq 3, \text{ simplify the expression:} \\&\qquad \frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \quad (\text{since } x \neq 3) \\&\text{So, the function can be rewritten as:} \\&\qquad f(x) = \begin{cases} x + 3, & x \neq 3 \\ 5, & x = 3 \end{cases} \\\\&\textbf{Compute the Limit as } x \to 3 \\&\text{Evaluate the limit of } f(x) \text{ as } x \text{ approaches } 3: \\&\qquad \lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 3 + 3 = 6 \\\\&\textbf{Compare the Limit and the Function Value at } x = 3 \\&\text{The function value at } x = 3 \text{ is given as:} \\&\qquad f(3) = 5 \\\\&\textbf{Check Continuity} \\&\text{For } f(x) \text{ to be continuous at } x = 3, \text{ the following must hold:} \\&\qquad \lim_{x \to 3} f(x) = f(3) \\&\text{However, we have:} \\&\qquad 6 \neq 5 \\&\text{Since the limit does not equal the function value at } x = 3, \text{ the function is not continuous at } x = 3.\end{aligned}$$​​