Question

If $f(x) = \begin{cases} \dfrac{x^{2} - 9}{x - 3}, & x \neq 3 \\ 2x + k, & x = 3 \end{cases}$ is continuous at $x = 3$, then $k$ equals

• A. 3
• B. 0
• C. $-6$
• D. $\dfrac{1}{6}$

Hint:

Factor and cancel first: $\dfrac{x^2-9}{x-3} = x+3$ for $x \neq 3$. The limit at $x=3$ is $6$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For continuity: $\lim_{x\to3}f(x) = f(3)$.
Step 2: Detailed Explanation:
$\lim_{x\to3}\dfrac{x^2-9}{x-3} = \lim_{x\to3}(x+3) = 6$.
$f(3) = 2(3)+k = 6+k$.
Continuity: $6+k = 6 \Rightarrow k = 0$.
Step 3: Final Answer:
$k = 0$.