If the function \(f(x) = \begin{cases}\displaystyle\frac{x^{3} - 27}{x - 3}, & x \ne 3\\ k, & x = 3\end{cases}\) is continuous at \(x = 3\), then the value of k is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 27
f is continuous at 3 when \(f(3) = \lim_{x \to 3} f(x)\), so k equals this limit.
Using the difference of cubes \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\):
\( \displaystyle \frac{x^{3} - 27}{x - 3} = \frac{(x - 3)(x^{2} + 3x + 9)}{x - 3} \)
\( = x^{2} + 3x + 9, \quad x \ne 3 \)
\( \displaystyle \lim_{x \to 3}(x^{2} + 3x + 9) = 9 + 9 + 9 = 27 \)
So k = 27.