1. \(=\lim\limits_{x\rightarrow3}\dfrac{\left(3-x\right)\left(3+x\right)\left(\sqrt{x+6}+3\right)}{x-3}=\lim\limits_{x\rightarrow3}-\left(3+x\right)\left(\sqrt{x+6}+3\right)=-36\)
a) \(lim_{x\rightarrow3}\dfrac{9-x^2}{\sqrt{x+6}-3}=lim_{x\rightarrow3}\dfrac{\left(\sqrt{x+6}+3\right)\left(3-x\right)\left(3+x\right)}{\left(\sqrt{x+6}-3\right)\left(\sqrt{x+6}+3\right)}\)
\(=lim_{x\rightarrow3}\dfrac{\left(\sqrt{x+6}+3\right)\left(x-3\right)\left(-x-3\right)}{x+6-9}\)
\(=lim_{x\rightarrow3}\left(\sqrt{x+6}+3\right)\left(-x-3\right)\)
\(=\left(\sqrt{3+6}+3\right)\left(-3-3\right)=-36\)
b) \(lim_{x\rightarrow-\infty}\left(\sqrt{x^2+x}+x\right)=lim_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+x}+x\right)\left(\sqrt{x^2+x}-x\right)}{\sqrt{x^2+x}-x}\)
\(=lim_{x\rightarrow-\infty}\dfrac{x}{\sqrt{x^2+x}-x}=lim_{x\rightarrow-\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x}}-1}=\dfrac{-1}{2}\)
