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

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

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

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

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

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

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

Không tồn tại.

\(a=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2-2x-2\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2-2x-2}{x-3}=\dfrac{3}{2}\)

Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)

\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)

 \(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)

a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)

Từ 2 câu kia lát tui làm, ăn cơm đã :D

1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)

2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)

3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)

Xai L'Hospital nhe :v

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-2+\sqrt{5x+4}-3}{x-1}\)

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

Đề câu b là \(...\sqrt{90-6x}\) hay \(\sqrt{9-6x}\) vậy em? Hình như cái sau mới có lý

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

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

b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)

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

a: \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)

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

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)\cdot\left(\sqrt{3x+1}+2\right)}{3\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}{3}\)

\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)

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

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

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

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

\(=2\cdot\left(\sqrt{2+2}+2\right)\)

\(=2\cdot4=8\)