For what value of $k$, the function given below is continuous at $x = 0$?

$f(x) = \begin{cases} \frac{\sqrt{4 + x} - 2}{x} &, x \neq 0 \\ k &, x = 0 \end{cases}$

$\frac{1}{4}$

The correct answer is Option (2) → $\frac{1}{4}$ ##

$f(x) = \begin{cases} \frac{\sqrt{4 + x} - 2}{x} &, x \neq 0 \\ k &, x = 0 \end{cases}$

For continuity at $x = 0$:

$\lim\limits_{x \to 0} [f(x)] = \lim\limits_{x \to 0^+} [f(x)] = f(0)$

Rationalizing the numerator:

$f(x) = \frac{\sqrt{4 + x} - 2}{x} \cdot \frac{\sqrt{4 + x} + 2}{\sqrt{4 + x} + 2}$

$f(x) = \frac{4 + x - 4}{x(\sqrt{4 + x} + 2)}$

$f(x) = \frac{1}{\sqrt{4 + x} + 2}$

Evaluating the limit:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{1}{\sqrt{4 + 0} + 2} = \frac{1}{4}$

Given $f(0) = k$.

So, $k = \frac{1}{4}$