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

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

\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{4x+5-4x^2-12x-9}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)

\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4x^2-8x-4}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)

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

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

\(=\dfrac{-4}{\sqrt{-4+5}-2+3}=\dfrac{-4}{1+1}=-\dfrac{4}{2}=-2\)

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

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

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

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

=0

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

Không tồn tại.

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

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

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

$x\to -2$ thì $(x+2)^2\to 0$

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

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

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

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?

1.

\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)

\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:

\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)

2.

\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:

\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)

3.

\(\lim\limits_{x\to +\infty}\frac{3x^2-5\sin 2x+7\cos ^2x}{2x^2+2}=\lim\limits_{x\to +\infty}\frac{3x^2-5\sin 2x+7(1-\sin ^2x)}{2x^2+2}\)

\(=\lim\limits_{x\to +\infty}\frac{3(x^2+1)-5\sin 2x+4-7\sin ^2x}{2x^2+2}\)

\(=\lim\limits_{x\to +\infty}\left[\frac{3}{2}-5.\frac{\sin 2x}{2x}.\frac{2x}{2x^2+2}+\frac{2}{x^2+1}-7.(\frac{\sin x}{x})^2.\frac{x^2}{2x^2+2}\right]\)

\(=\frac{3}{2}-5.0.0+0-7.0.\frac{1}{2}=\frac{3}{2}\) (nhớ rằng \(\lim\limits_{t\to \infty}\frac{\sin t}{t}=0\))