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

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

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

Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

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

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)

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

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

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

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)

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

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

\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)

Check lai ho minh nhe :v

Hic nan qua :( Lam vay

P/s: Anh Lam check all ho em nhung bai em lam nhe :( Em cam on

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

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

3/ \(=\lim\limits_{x\rightarrow-\infty}x^5\left(4-\dfrac{3}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^5}\right)=-\infty\)

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

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

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

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

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}{\dfrac{\left(\sqrt[3]{x^3+x^2+1}\right)^2}{x^2}+\dfrac{x}{x^2}\sqrt[3]{x^3+x^2+1}+\dfrac{x^2}{x^2}}+\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}\)

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

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

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

5/ \(=\lim\limits_{x\rightarrow+\infty}x.\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\right)=+\infty\)