The value of \(k\) for which the function, defined by, \(f(x) = \begin{cases} \displaystyle\frac{3x + 4\tan x}{x} & : x \ne 0 \\ k & : x = 0 \end{cases}\) is continuous at \(x = 0\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7
f is continuous at x = 0 when \(\lim_{x \to 0} f(x) = f(0) = k\).
\( \displaystyle \lim_{x \to 0} \frac{3x + 4\tan x}{x} = \lim_{x \to 0} \left(3 + 4 \cdot \frac{\tan x}{x}\right) \)
Using the standard limit \(\lim_{x \to 0} \frac{\tan x}{x} = 1\):
\(= 3 + 4(1) = 7\)
So \(k = 7\).