\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)

= hệ số x^n-1 trên tử/hệ số x^n-1 dưới mẫu  = \(\dfrac{a_1+a_2+...+a_n}{n}\)

a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x+1\right)^2-\left(x^2-x+2\right)}{x+1+\sqrt{x^2-x+2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{3x-1}{x+1+\sqrt{x^2-x+2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{3-\dfrac{1}{x}}{1+\dfrac{1}{x}+\sqrt{1-\dfrac{1}{x}+\dfrac{2}{x^2}}}=\dfrac{3}{2}\)

\(\lim\limits_{x\rightarrow+\infty}=\left(\sqrt[3]{x^3+3\text{x}^2}-\sqrt{x^2-2\text{x}}\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+3\text{x}^2}-x+x-\sqrt{x^2-2x}\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{3\text{x}^2}{\sqrt[3]{\left(x^3+3\text{x}^2\right)^2}+x\sqrt[3]{x^3+3\text{x}^2}+x^2}+\dfrac{2\text{x}}{x+\sqrt{x^2-2x}}\right)\\ =\dfrac{3}{1+1+1}+\dfrac{2}{1+1}=2\)

a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-2}{3-\sqrt{x^2+7}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}}{\dfrac{3}{x}-\sqrt{1+\dfrac{7}{x^2}}}\)

\(=\dfrac{1}{0-\sqrt{1+0}}=\dfrac{1}{-1}=-1\)

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x}-\sqrt{4x^2+1}}{2x+3}\)

\(=\dfrac{\sqrt{x^2\left(1-\dfrac{1}{x}\right)}-\sqrt{x^2\left(4+\dfrac{1}{x^2}\right)}}{2x+3}\)

\(=\dfrac{-x\cdot\sqrt{1-\dfrac{1}{x}}+x\cdot\sqrt{4+\dfrac{1}{x^2}}}{x\left(2+\dfrac{3}{x}\right)}\)

\(=\dfrac{-\sqrt{1-\dfrac{1}{x}}+\sqrt{4+\dfrac{1}{x^2}}}{2+\dfrac{3}{x}}=\dfrac{-1+2}{2}=\dfrac{1}{2}\)

Sao anh không thấy đề cụ thể ta!

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{\dfrac{4x^2+2x-1}{x^2}}-\dfrac{x}{x}}{\dfrac{3x-2}{x}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}-1}{3-\dfrac{2}{x}}=-\dfrac{4-1}{3}=-1\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}+1}{-3+\dfrac{2}{x}}=\dfrac{\sqrt{4}+1}{-3}=-1\).