#### Answer

$-\dfrac{3}{2}$

#### Work Step by Step

We want to find $\lim\limits_{x \to -2}\dfrac{x+2}{\sqrt{x^2+5}-3}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -2 is zero, and if we try to substitute $x=-2$ directly, we get zero in both the numerator and the denominator.
But note that $\dfrac{x+2}{\sqrt{x^2+5}-3}=\left(\dfrac{x+2}{\sqrt{x^2+5}-3}\right)\left(\dfrac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3}\right)=\dfrac{(x+2)(\sqrt{x^2+5}+3)}{x^2+5-9}=\dfrac{(x+2)(\sqrt{x^2+5}+3)}{x^2-4}=\dfrac{(x+2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)}=\dfrac{\sqrt{x^2+5}+3}{x-2}$ for all $x \neq -2.$
Thus $\lim\limits_{x \to -2}\dfrac{x+2}{\sqrt{x^2+5}-3}=\lim\limits_{x \to -2}\dfrac{\sqrt{x^2+5}+3}{x-2}=\dfrac{\sqrt{4+5}+3}{-2-2}=\dfrac{\sqrt{9}+3}{-4}=\dfrac{3+3}{-4}=\dfrac{6}{-4}=-\dfrac{3}{2}.$