If $f(x)=\left\{\begin{matrix}\frac{\sqrt{4+x}-2}{x}, & if\, x≠0\\k ,& if \, x ≠0\end{matrix}\right.$ is continuous at x=0, then the value of k is :

$\frac{1}{4}$

The correct answer is Option (3) → $\frac{1}{4}$

$f(0)=k$

$\underset{x→0}{\lim}\frac{\sqrt{4+x}-2}{x}\frac{(\sqrt{4+x}+2)}{(\sqrt{4+x}+2)}=\underset{x→0}{\lim}\frac{x}{x(\sqrt{4+x}+2)}$

$=\underset{x→0}{\lim}\frac{1}{\sqrt{4+x}+2}=\frac{1}{4}=k$