\(\lim\limits_{x\rightarrow0}\dfrac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\dfrac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}\dfrac{4\left(\sqrt{9+x}+3\right)}{1}=4\left(\sqrt{9}+3\right)=24\)
\(\lim\limits_{x\rightarrow1}\dfrac{3x-2\sqrt{4x^2-x-2}}{x^2-3x+2}=\dfrac{3-2\sqrt{4-1-2}}{0}=\dfrac{1}{0}=+\infty\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{2x+7}+3}+x-1}{\left(x-1\right)\left(x^2-3x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2}{\sqrt{2x+7}+3}+1}{x^2-3x-3}=\dfrac{\dfrac{2}{\sqrt{9}+3}+1}{1-3-3}=-\dfrac{4}{15}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt{x+1}+1}+\dfrac{x}{\sqrt{x+4}+2}}{x}=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{x+4}+2}\right)=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{4x}-2}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(\sqrt[3]{\left(4x\right)^2}+2\sqrt[3]{4x}+4\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\sqrt[3]{\left(4x\right)^2}+2\sqrt[3]{4x}+4}=\dfrac{4}{\sqrt[3]{64}+2\sqrt[3]{8}+4}=...\)
