\(\lim\limits_{x\rightarrow4}\frac{2x-\sqrt{3x+1}}{x^2-1}=\frac{8-\sqrt{11}}{15}\)
Nhưng mình đoán bạn ghi nhầm đề, x tiến tới 1 mới có lý
\(\lim\limits_{x\rightarrow1}\frac{2x-\sqrt{3x+1}}{x^2-1}=\lim\limits_{x\rightarrow1}\frac{4x^2-3x-1}{\left(x-1\right)\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(4x+1\right)}{\left(x-1\right)\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}=\lim\limits_{x\rightarrow1}\frac{4x+1}{\left(x+1\right)\left(2x+\sqrt{3x+1}\right)}=\frac{5}{2\left(2+2\right)}=\frac{5}{8}\)
\(\lim\limits_{x\rightarrow8}\frac{\sqrt[3]{x}-2+2-\sqrt{x-4}}{x-8}=\lim\limits_{x\rightarrow8}\frac{\frac{x-8}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}-\frac{x-8}{2+\sqrt{x-4}}}{x-8}\)
\(=\lim\limits_{x\rightarrow8}\left(\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}-\frac{1}{2+\sqrt{x-4}}\right)=\frac{1}{12}-\frac{1}{4}=-\frac{1}{6}\)