Chương 4: GIỚI HẠN

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Shan darren

 

xin mọi người giúp em vói ạundefined

Nguyễn Việt Lâm
11 tháng 3 2022 lúc 0:11

13.

\(\lim\limits_{x\rightarrow1}\left(3x^2-2x-1\right)=3.1^2-2.1-1=0\)

14.

\(\lim\limits_{x\rightarrow9}\sqrt{x+16}=\sqrt{9+16}=5\)

15.

\(\lim\limits_{x\rightarrow+\infty}x^{2021}=+\infty\)

16.

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

Do \(\lim\limits_{x\rightarrow-\infty}x^3=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=-1< 0\)

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

Nguyễn Việt Lâm
11 tháng 3 2022 lúc 0:15

17.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+5x-3}{x^2+6x+3}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{5}{x}-\dfrac{3}{x^2}\right)}{x^2\left(1+\dfrac{6}{x}+\dfrac{3}{x^2}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{5}{x}-\dfrac{3}{x^2}}{1+\dfrac{6}{x}+\dfrac{3}{x^2}}=\dfrac{2+0-0}{1+0+0}=2\)

18.

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

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

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

Nguyễn Việt Lâm
11 tháng 3 2022 lúc 0:18

19.

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

Do \(\lim\limits_{x\rightarrow-\infty}x^3=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\left(2-\dfrac{1}{x}\right)=2>0\)

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

20.

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

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

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{0}{1+1}=0\)

Nguyễn Việt Lâm
11 tháng 3 2022 lúc 0:21

21.

\(\lim\limits_{x\rightarrow2}\dfrac{x^3-8}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4}{x+2}=\dfrac{2^2+2.2+4}{2+2}=\dfrac{12}{4}=3\)

22.

\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{x^2-x-2}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)\left(x-2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(x+1\right)=2+1=3\)

\(f\left(2\right)=m\)

Hàm liên tục tại \(x=2\) khi \(f\left(2\right)=\lim\limits_{x\rightarrow2}f\left(x\right)\Rightarrow m=3\)

Nguyễn Việt Lâm
11 tháng 3 2022 lúc 0:23

23.

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

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

\(f\left(1\right)=3.1+m=m+3\)

Hàm liên tục tại \(x=1\) khi:

\(\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow m+3=3\)

\(\Rightarrow m=0\)


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