\(\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{\sqrt{8+2x}-2}{\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2x+8-4}{\left(\sqrt{2x+8}+2\right)\cdot\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2\cdot\sqrt{x+2}}{\sqrt{2x+8}+2}=\dfrac{2\cdot\sqrt{-2+2}}{\sqrt{2\cdot\left(-2\right)+8}+2}\)
=0