Tính: \(lim_{x\rightarrow-2}\dfrac{2x+1}{\left(x+2\right)^2}\)
K
Khách
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TK
1
24 tháng 11 2023
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}-2x-3}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{4x+5-\left(2x+3\right)^2}{\sqrt{4x+5}+2x+3}\cdot\dfrac{1}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{4x+5-4x^2-12x-9}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4x^2-8x-4}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4\left(x^2+2x+1\right)}{\left(x+1\right)^2\cdot\left(\sqrt{4x+5}+2x+3\right)}\right)\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{-4}{\sqrt{4x+5}+2x+3}\)
\(=\dfrac{-4}{\sqrt{-4+5}-2+3}=\dfrac{-4}{1+1}=-\dfrac{4}{2}=-2\)
C
1
23 tháng 12 2023
\(\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{\sqrt{8+2x}-2}{\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2x+8-4}{\left(\sqrt{2x+8}+2\right)\cdot\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2\cdot\sqrt{x+2}}{\sqrt{2x+8}+2}=\dfrac{2\cdot\sqrt{-2+2}}{\sqrt{2\cdot\left(-2\right)+8}+2}\)
=0
23 tháng 12 2023
\(\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\cdot\sqrt{\dfrac{3x}{x^2-1}}\)
\(=\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^2-x+1\right)\left(x+1\right)\cdot\dfrac{\sqrt{3x}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow\left(-1\right)^-}\sqrt{x+1}\cdot\left(x^2-x+1\right)\cdot\sqrt{\dfrac{3x}{x-1}}\)
\(=\sqrt{-1+1}\left[\left(-1\right)^2-\left(-1\right)+1\right]\cdot\sqrt{\dfrac{3\left(-1\right)}{-1-2}}\)
=0
HT
9 tháng 2 2021
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 3 2021
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1+1-\sqrt[]{1-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{x}{1+\sqrt[]{1-x}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[3]{\left(x+1\right)^3}+\sqrt[3]{x+1}+1}+\dfrac{1}{1+\sqrt[]{1-x}}\right)=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)
AH
Akai Haruma
Giáo viên
27 tháng 11 2023
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 $x-1<0$
\(\lim\limits_{x\to 6}\frac{(5x-4)\sqrt{2x-3}+x-84}{x-6}=\lim\limits_{x\to 6}\frac{(5x-4)(\sqrt{2x-3}-3)+16(x-6)}{x-6}\)
\(=\lim\limits_{x\to 6}\frac{(5x-4).\frac{2(x-6)}{\sqrt{2x-3}+3}+16(x-6)}{x-6}=\lim\limits_{x\to 6}[\frac{2(5x-4)}{\sqrt{2x-3}+3}+16]=\frac{74}{3}\)
Lời giải:
$x\to -2$ thì $2x+1\to -3<0$
$x\to -2$ thì $(x+2)^2\to 0$
$\Rightarrow \lim\limits_{x\to -2}\frac{2x+1}{(x+2)^2}=-\infty$