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

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

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

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

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

=0

Lời giải:
$x\to -2$ thì $2x+1\to -3<0$

$x\to -2$ thì $(x+2)^2\to 0$

$\Rightarrow \lim\limits_{x\to -2}\frac{2x+1}{(x+2)^2}=-\infty$

\(\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{\sqrt{8+2x}-2}{\sqrt{x+2}}\)

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

\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2\cdot\sqrt{x+2}}{\sqrt{2x+8}+2}=\dfrac{2\cdot\sqrt{-2+2}}{\sqrt{2\cdot\left(-2\right)+8}+2}\)

=0

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

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[3]{\left(x+1\right)^3}+\sqrt[3]{x+1}+1}+\dfrac{1}{1+\sqrt[]{1-x}}\right)=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)

Chọn \(f\left(x\right)=5x+5\)

Khi đó: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)

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

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

p/s: bài này mình chưa học trên lớp nên ko chắc 100% đúng

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

Câu c số 1 trong hay ngoài căn nhỉ?

Lời giải:

\(\lim\limits_{x\to 1-}\frac{2x+1}{x-1}=-\infty\) do với $x\to 1-$ thì $\lim(2x+1)=3>0$ và $\lim (x-1)=0$ và $x-1<0$

\(\lim\limits_{x\to 6}\frac{(5x-4)\sqrt{2x-3}+x-84}{x-6}=\lim\limits_{x\to 6}\frac{(5x-4)(\sqrt{2x-3}-3)+16(x-6)}{x-6}\)

\(=\lim\limits_{x\to 6}\frac{(5x-4).\frac{2(x-6)}{\sqrt{2x-3}+3}+16(x-6)}{x-6}=\lim\limits_{x\to 6}[\frac{2(5x-4)}{\sqrt{2x-3}+3}+16]=\frac{74}{3}\)