If the function \(f(x) = \begin{cases}\displaystyle\frac{\log_e(1 + 5x)}{x}, & x \ne 0\\ k, & x = 0\end{cases}\), where \(x \gt -\displaystyle\frac{1}{5}\), is continuous at \(x = 0\), then the value of k is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
For continuity at 0, k must equal the limit of f at 0. Using the standard limit \(\lim_{t \to 0}\frac{\log_e(1 + t)}{t} = 1\) with \(t = 5x\):
\( \displaystyle \frac{\log_e(1 + 5x)}{x} = 5 \cdot \frac{\log_e(1 + 5x)}{5x} \)
\( \displaystyle \lim_{x \to 0} f(x) = 5 \times 1 = 5 \)
So k = 5.