#### Answer

$\dfrac{3}{2}$

#### Work Step by Step

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