For a function to be continuous at a point, the limit of the function as it approaches that point must exist and be equal to the function's value at that point. In this case, we need to find the limit of $f(x)$ as $x$ approaches 0 and set it equal to $k$.
We need to find $\lim_{x \to 0} \frac{\sqrt{4+x}-2}{x}$. This limit is of the indeterminate form $\frac{0}{0}$, so we can use L'Hôpital's rule or rationalize the numerator.
Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{4+x}+2$:\r\n$$\lim_{x \to 0} \frac{\sqrt{4+x}-2}{x} \cdot \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2} = \lim_{x \to 0} \frac{(4+x)-4}{x(\sqrt{4+x}+2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{4+x}+2)}$$
Cancel out the $x$ in the numerator and denominator:\r\n$$\lim_{x \to 0} \frac{1}{\sqrt{4+x}+2}$$
Now, substitute $x=0$ into the simplified expression:\r\n$$\frac{1}{\sqrt{4+0}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}$$
For the function to be continuous at $x=0$, we must have $f(0) = \lim_{x \to 0} f(x)$. Therefore, $k = \frac{1}{4}$.
Final Answer: 1/4
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