If $f(x)=\left\{\begin{matrix} \frac{\sqrt{x^2+5}-3}{x+2} &, & x≠-2\\k & , & x=-2 \end{matrix}\right.$ is continuous at x=-2 then the value of k is :

$-\frac{2}{3}$

The correct answer is option (1) → $-\frac{2}{3}$

$f(-2)=k$

$\lim\limits_{x→-2}\frac{\sqrt{x^2+5}-3}{x+2}=\frac{0}{0}form$

Using L'Hopital's rule

$\lim\limits_{x→-2}\frac{\frac{1}{2}\frac{(2x)}{\sqrt{x^2+5}}}{1}=\lim\limits_{x→-2}\frac{x}{\sqrt{x^2+5}}=-\frac{2}{\sqrt{9}}=-\frac{2}{3}$