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NL
6 tháng 3 2022 lúc 21:44

a.

\(\lim\limits_{x\rightarrow2}\dfrac{x\sqrt{2x}+\sqrt{2x}-6}{x^2+2x-8}=\lim\limits_{x\rightarrow2}\dfrac{\left(x\sqrt{2x}-4\right)+\left(\sqrt{2x}-2\right)}{\left(x-2\right)\left(x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{2x^3-16}{x\sqrt{2x}+4}+\dfrac{2x-4}{\sqrt{2x}+2}}{\left(x-2\right)\left(x+4\right)}\)

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

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

\(=\dfrac{\dfrac{2\left(2^2+2.2+4\right)}{2\sqrt{4}+4}+\dfrac{2}{\sqrt{4}+2}}{2+4}\)

\(=...\)

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NL
6 tháng 3 2022 lúc 21:46

b.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{9x^2-3x+2}-3}{x+2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left|x\right|\sqrt{9-\dfrac{3}{x}+\dfrac{2}{x^2}}-3}{x+2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{9-\dfrac{3}{x}+\dfrac{2}{x^2}}-3}{x+2}\)

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

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{9-\dfrac{3}{x}+\dfrac{2}{x^2}}-\dfrac{3}{x}}{1+\dfrac{2}{x}}\)

\(=\dfrac{-\sqrt{9-0+0}-0}{1+0}=...\)

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NL
6 tháng 3 2022 lúc 21:48

c.

\(\lim\limits_{x\rightarrow1^-}\dfrac{1-2x^2}{x-1}=\lim\limits_{x\rightarrow1^-}\dfrac{2x^2-1}{1-x}\)

Ta có:

\(\lim\limits_{x\rightarrow1^-}\left(2x^2-1\right)=1>0\)

\(\lim\limits_{x\rightarrow1^-}\left(1-x\right)=0\)

Và \(1-x>0\) với mọi \(x< 1\)

\(\Rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{2x^2-1}{1-x}=+\infty\)

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NL
6 tháng 3 2022 lúc 21:50

d.

\(\lim\limits_{x\rightarrow3}\dfrac{x-\sqrt{2x+3}}{x^2-3x}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-\sqrt{2x+3}\right)\left(x+\sqrt{2x+3}\right)}{\left(x^2-3x\right)\left(x+\sqrt{2x+3}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x^2-2x-3}{x\left(x-3\right)\left(x+\sqrt{2x+3}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x-3\right)\left(x+\sqrt{2x+3}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x+1}{x\left(x+\sqrt{2x+3}\right)}\)

\(=\dfrac{3+1}{3\left(3+\sqrt{9}\right)}=...\)

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