If the function \(f(x) = \begin{cases}\displaystyle\frac{e^{6x} - 1}{\sin 3x}, & x \ne 0\\ k, & x = 0\end{cases}\) is continuous at \(x = 0\), then the value of k is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
For continuity at 0, k must equal \(\lim_{x \to 0} f(x)\).
Using the standard limits \(\lim_{t \to 0}\frac{e^{t} - 1}{t} = 1\) and \(\lim_{t \to 0}\frac{\sin t}{t} = 1\), write
\( \displaystyle \frac{e^{6x} - 1}{\sin 3x} = \frac{e^{6x} - 1}{6x} \times \frac{3x}{\sin 3x} \times \frac{6x}{3x} \)
As \(x \to 0\), the first two factors tend to 1 and the last factor equals 2:
\( \displaystyle \lim_{x \to 0}\frac{e^{6x} - 1}{\sin 3x} = 1 \times 1 \times 2 = 2 \)
So k = 2.