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

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)

Da nan roi mang meo lam mat het bai -.-

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

2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)

3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)

1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)

Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5

2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)

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

4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)

5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)

6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)

1/ \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}+\dfrac{x}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)=\dfrac{13}{12}\)

2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2-\left(\sqrt{x+3}-2\right)}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{x-1}{\sqrt{x+3}+2}}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{1}{\sqrt{x+3}+2}}{x-2}=\dfrac{1}{6}\)

3/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-2+2-\sqrt{5-x^2}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(x^2-1\right)}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\left(\dfrac{1}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{1}{2+\sqrt{5-x^2}}\right)=\dfrac{1}{3}\)

4/ \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}=\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-3-\left(\sqrt[3]{8x+43}-3\right)}{\left(2x-1\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{x+2}{\sqrt{x+11}+3}-\dfrac{8\left(x+2\right)}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{\left(2x-1\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{1}{\sqrt{x+11}+3}-\dfrac{8}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{2x-1}=\dfrac{7}{270}\)

5/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1-\left(\sqrt[m]{1+bx}-1\right)}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{bx}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{b}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)

\(=\dfrac{a}{n}-\dfrac{b}{m}\)

6/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-\sqrt{1+4x}+\sqrt{1+4x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\left(\sqrt[3]{1+6x}-1\right)+\sqrt{1+4x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\dfrac{6x}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4x}{\sqrt{1+4x}+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{6\sqrt{1+4x}}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4}{\sqrt{1+4x}+1}\right)=4\)

1/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7-9\right)\left(2+\sqrt{x+3}\right)}{\left(4-x-3\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(2+\sqrt{x+3}\right)}{\left(x-1\right)\left(-\sqrt{2x+7}-3\right)}=\dfrac{2.4}{-6}=-\dfrac{4}{3}\)

2/ \(=\lim\limits_{x\rightarrow1^-}\dfrac{2.1-3}{1-1}=-\infty\)

3/ \(=\lim\limits_{x\rightarrow2^+}\dfrac{3-x}{x-2}=+\infty\)

4/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-\dfrac{8x^3}{x^2}+\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}-\dfrac{1}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-8x}{5}=\pm\infty\)

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

a: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)

\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)

\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)

c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)

\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)

\(=-\sqrt{2+2}-2=-2-2=-4\)

\(a=\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x^3-3x-2\right)\left(x+\sqrt[]{x+2}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+1\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x-2\right)\left(x+1\right)^2\left(x+\sqrt[]{x+2}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}}{\left(x+1\right)\left(x+\sqrt[]{x+2}\right)}=...\)

\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+2x}-x-1\right)+\left(x+1-\sqrt[3]{1+3x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x^2}{\sqrt[]{1+2x}+x+1}+\dfrac{x^3+3x^2}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[]{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)\)

\(=...\)

\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{5+4x}-2x-3\right)+\left(2x+3-\sqrt[3]{7+6x}\right)}{x^3+x^2-x-1}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{5+4x-\left(2x+3\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(2x+3\right)^3-\left(7+6x\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4\left(x+1\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(x+1\right)^2\left(8x+20\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4}{2x+3+\sqrt[]{5+4x}}+\dfrac{8x+20}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{x-1}\)

\(=...\)

Hiển nhiên là cách đầu sai rồi em

Khi đến \(\lim x^2\left(1-1\right)=+\infty.0\) là 1 dạng vô định khác, đâu thể kết luận nó bằng 0 được

a/ L'Hospital:

 \(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)

b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)