Question:medium
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
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
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Factor and cancel first: $\dfrac{x^2-9}{x-3} = x+3$ for $x \neq 3$. The limit at $x=3$ is $6$.
Updated On: Apr 8, 2026
- 3
- 0
- $-6$
- $\dfrac{1}{6}$
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The Correct Option is B
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