Lời giải:
Có: \(\lim_{x\to +\infty}\frac{2\sqrt{x}-3\sqrt{x+3}+2}{x-1}=\lim_{x\to +\infty}\frac{\frac{2}{\sqrt{x}}-3\sqrt{\frac{x+3}{x^2}}+\frac{2}{x}}{1-\frac{1}{x}}\)
Mà:
\(\lim_{x\to +\infty}(\frac{2}{\sqrt{x}}-3\sqrt{\frac{x+3}{x^2}}+\frac{2}{x})=\lim_{x\to +\infty}(\frac{2}{\sqrt{x}}-3\sqrt{\frac{1}{x}+\frac{3}{x^2}}+\frac{2}{x})=0-0+0=0\)
\(\lim_{x\to +\infty}(1-\frac{1}{x})=1-\lim _{x\to +\infty}\frac{1}{x}=1-0=1\)
\(\Rightarrow \lim_{x\to +\infty}\frac{2\sqrt{x}-3\sqrt{x+3}+2}{x-1}=\frac{0}{1}=0\)