1. lim (\(\dfrac{1}{x}+\dfrac{1}{x^2}\) ) khi x->0
2 lim \(\dfrac{x^3+8}{X+2}\) khi x->-2
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MH
3 tháng 2 2021
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
TL
7 tháng 11 2023
Em là tám lại ạ
Em là duy khôi ạ
Em là văn tam ạ
Em là mạnh Tuấn ạ
8 tháng 11 2023
a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}\)
\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)
b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x-3+x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2\)
d: \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x+x-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\dfrac{x^3-x^3+1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{x^2+1}+x}+\dfrac{1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}+\dfrac{\dfrac{1}{x^2}}{\sqrt[3]{\dfrac{1}{x^4}}+\sqrt[3]{1-\dfrac{1}{x^3}}+\sqrt[3]{\left(1-\dfrac{1}{x^3}\right)^2}}\right)\)
=0
c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(a=\lim\limits_{x\rightarrow1^+}\dfrac{x^2-x+1}{x^2-1}=\dfrac{1}{0}=+\infty\)
\(b=\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8+x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}-\dfrac{x}{4+2\sqrt[3]{8+x}+\sqrt[3]{\left(8+x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}-\dfrac{1}{4+2\sqrt[3]{8+x}+\sqrt[3]{\left(8+x\right)^2}}\right)=\dfrac{2}{2}-\dfrac{1}{12}=...\)
\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(\sqrt{2x-2}+2\right)}{2\left(x-3\right)\left(\sqrt{x+6}+3\right)}=\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x-2}+2}{2\left(\sqrt{x+6}+3\right)}=\dfrac{2+2}{2\left(3+3\right)}=...\)
HT
16 tháng 4 2021
1/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7-9\right)\left(2+\sqrt{x+3}\right)}{\left(4-x-3\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(2+\sqrt{x+3}\right)}{\left(x-1\right)\left(-\sqrt{2x+7}-3\right)}=\dfrac{2.4}{-6}=-\dfrac{4}{3}\)
2/ \(=\lim\limits_{x\rightarrow1^-}\dfrac{2.1-3}{1-1}=-\infty\)
3/ \(=\lim\limits_{x\rightarrow2^+}\dfrac{3-x}{x-2}=+\infty\)
4/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-\dfrac{8x^3}{x^2}+\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}-\dfrac{1}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-8x}{5}=\pm\infty\)
5/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}}+\dfrac{2x}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\dfrac{7}{x}}=\dfrac{1}{2}\)
HT
9 tháng 2 2021
1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)
2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)
3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)
4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v
NT
1
AH
Akai Haruma
Giáo viên
26 tháng 1 2021
Lời giải:
\(\lim\limits_{x\to 0}\frac{2-\sqrt{4-x}}{x}=\lim\limits_{x\to 0}\frac{4-(4-x)}{x(2+\sqrt{4-x})}=\lim\limits_{x\to 0}\frac{x}{x(2+\sqrt{4-x})}=\lim\limits_{x\to 0}\frac{1}{2+\sqrt{4-x}}=\frac{1}{4}\)
\(\lim\limits_{x\to -2}\frac{x^3+8}{x^2-4}=\lim\limits_{x\to -2}\frac{(x+2)(x^2-2x+4)}{(x-2)(x+2)}=\lim\limits_{x\to -2}\frac{x^2-2x+4}{x-2}=-3\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2022
1.
Do \(\lim\limits_{x\rightarrow2}\left(3x-5\right)=1>0\)
\(\lim\limits_{x\rightarrow2}\left(x-2\right)^2=0\)
\(\left(x-2\right)^2>0;\forall x\ne2\)
\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{3x-5}{\left(x-2\right)^2}=+\infty\)
2.
\(\lim\limits_{x\rightarrow1^-}\left(2x-7\right)=-5< 0\)
\(\lim\limits_{x\rightarrow1^-}\left(x-1\right)=0\)
\(x-1< 0;\forall x< 1\)
\(\Rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{2x-7}{x-1}=+\infty\)
3.
\(\lim\limits_{x\rightarrow1^+}\left(2x-7\right)=-5< 0\)
\(\lim\limits_{x\rightarrow1^+}\left(x-1\right)=0\)
\(x-1>0;\forall x>1\)
\(\Rightarrow\lim\limits_{x\rightarrow1^+}\dfrac{2x-7}{x-1}=-\infty\)
\(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{x}+\dfrac{1}{x^2}\right)=\infty+\infty=\infty\)
\(\lim\limits_{x\rightarrow-2}\left(\dfrac{x^3+8}{x+2}\right)=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x+2}=\lim\limits_{x\rightarrow-2}\left(x^2-2x+4\right)=12\)