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\(\hept{\begin{cases}lim_{x\rightarrow3^+}\frac{\left|x-3\right|}{x-3}=lim_{x\rightarrow3^+}\frac{x-3}{x-3}=1\\lim_{x\rightarrow3^-}\frac{\left|x-3\right|}{x-3}=lim_{x\rightarrow3^-}\frac{-x+3}{x-3}=-1\end{cases}\Rightarrow lim_{x\rightarrow3^+}\frac{\left|x-3\right|}{x-3}\ne lim_{x\rightarrow3^-}\frac{\left|x-3\right|}{x-3}}\)
=> đpcm
1
Ta có limx→3+|x−3| /(x−3) =limx→3+(x−3)/(x−3) =1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|/x−3=limx→3−−(x−3)/x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+|x−3|/
ta có:
lim x→3+ |x-3|/x-3 = lim x→3+ x-3/x-3 = 1/1 = 1 (1)
lim x→3- |x-3|/x-3 = lim x→3- -(x-3)/x-3 = -1/1 = -1 (2)
vì (1) ≠ (2) nên không tồn tại giới hạn lim x→3 |x-3|/x-3
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Đúng(0)
Ta có \lim\limits_{x\rightarrow 3^+} \dfrac{|x-3|}{x-3} = \lim\limits_{x\rightarrow 3^+} \dfrac{x-3}{x-3} = 1.x→3+limx−3∣x−3∣=x→3+limx−3x−3=1.
Mà \lim\limits_{x\rightarrow 3^-} \dfrac{|x-3|}{x-3} = \lim\limits_{x\rightarrow 3^-} \dfrac{-(x-3)}{x-3} = -1.x→3−limx−3∣x−3∣=x→3−limx−3−(x−3)=−1.
Vì \lim\limits_{x\rightarrow 3^+}\dfrac{|x-3|}{x-3} \ne \lim\limits_{x\rightarrow 3^-}\dfrac{|x-3|}{x-3}x→3+limx−3∣x−3∣=x→3
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Ta có \lim\limits_{x\rightarrow 3^+} \dfrac{|x-3|}{x-3} = \lim\limits_{x\rightarrow 3^+} \dfrac{x-3}{x-3} = 1.x→3+limx−3∣x−3∣=x→3+limx−3x−3=1.
Mà \lim\limits_{x\rightarrow 3^-} \dfrac{|x-3|}{x-3} = \lim\limits_{x\rightarrow 3^-} \dfrac{-(x-3)}{x-3} = -1.x→3−limx−3∣x−3∣=x→3−limx−3−(x−3)=−1.
Vì \lim\limits_{x\rightarrow 3^+}\dfrac{|x-3|}{x-3} \ne \lim\limits_{x\rightarrow 3^-}\dfrac{|x-3|}{x-3}x→3+limx−3∣x−3∣=x→3
Đúng(0)
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Ta có \lim\limits_{x\rightarrow 3^+} \dfrac{|x-3|}{x-3} = \lim\limits_{x\rightarrow 3^+} \dfrac{x-3}{x-3} = 1.x→3+limx−3∣x−3∣=x→3+limx−3x−3=1.
Mà \lim\limits_{x\rightarrow 3^-} \dfrac{|x-3|}{x-3} = \lim\limits_{x\rightarrow 3^-} \dfrac{-(x-3)}{x-3} = -1.x→3−limx−3∣x−3∣=x→3−limx−3−(x−3)=−1.
Vì \lim\limits_{x\rightarrow 3^+}\dfrac{|x-3|}{x-3} \ne \lim\limits_{x\rightarrow 3^-}\dfrac{|x-3|}{x-3}x→3+limx−3∣x−3∣=x→3
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Đúng(0)
không tồn tại giới hạn
không tồn tại
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Đúng(0)
Ta có limx→2√x2−4x+4x−2=limx→2|x−2|x−2limx→2x2−4x+4x−2=limx→2|x−2|x−2.
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Đúng(0)
lim không tồn tại giới hạn
Ta có \(\lim\limits_{x\rightarrow3^+}\dfrac{\left|x-3\right|}{x-3}=\lim\limits_{x\rightarrow3^+}\dfrac{x-3}{x-3}=1\)
Mà \(\lim\limits_{x\rightarrow3^-}\dfrac{\left|x-3\right|}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-\left(x-3\right)}{x-3}=-1\)
Vì \(\lim\limits_{x\rightarrow3^+}\dfrac{\left|x-3\right|}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{\left|x-3\right|}{x-3}\)
limx→3+|x−3|x−3≠limx→3−|x−3|x−3 nên không tồn tại giới hạn.
Ta có limx→3+|x−3|x−3=limx→3+x−3x−3=1.limx→3+|x−3|x−3=limx→3+x−3x−3=1.
Mà limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.limx→3−|x−3|x−3=limx→3−−(x−3)x−3=−1.
Vì limx→3+
Đúng(0)
Ta có: \(\lim\limits_{x\rightarrow3^+}\dfrac{\left|x-3\right|}{x-3}=\lim\limits_{x\rightarrow3^+}\dfrac{x-3}{x-3}=1.\)
Mà \(\lim\limits_{x\rightarrow3^-}\dfrac{\left|x-3\right|}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{\left|x-3\right|}{x-3}=-1.\)
Vì \(\lim\limits_{x\rightarrow3^+}\dfrac{\left|x-3\right|}{x-3}\ne\lim\limits_{x\rightarrow3^-}\dfrac{\left|x-3\right|}{x-3}\) nên không tồn tại giới hạn.
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-2}+\sqrt[3]{x^3+1}}{\sqrt{x^2+1}-x}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x+3}{\sqrt{2x^2-3}}\)
\(\lim\limits_{x\rightarrow\pm\infty}\dfrac{2x^2-1}{3-x^2}\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-x\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4}-1}{-1-1}=\dfrac{3}{2}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2x}{x}+\dfrac{3}{x}}{-\sqrt{\dfrac{2x^2}{x^2}-\dfrac{3}{x^2}}}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)
c/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)
Tính các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-3}\dfrac{x+3}{x^2+2x-3}\)
b) \(\lim\limits_{x\rightarrow0}\dfrac{\left(1+x\right)^3-1}{x}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-1}{x^2-1}\)
d) \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)
e) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-5}{\sqrt{x}+\sqrt{5}}\)
f) \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x^2+5}-3}{x+2}\)
g) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x}-1}{\sqrt{x+3}-2}\)
h) \(\lim\limits_{x\rightarrow+\infty}\dfrac{1-2x+3x^3}{x^3-9}\)
i) \(\lim\limits_{x\rightarrow0}\dfrac{1}{x^2}\left(\dfrac{1}{x^2+1}-1\right)\)
j) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^2+x^3+5}{4x^3+1}\)
b) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]` `ĐK: 4x^3+1 ne 0`
`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`
`=1/4`
`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]` `ĐK: x ne 0;x ne 1`
`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`
`=0`
Câu `c` giống `b`.
Tính các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2-1}{x+1}\)
b) \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{x+2}\)
c) \(\lim\limits_{x\rightarrow6}\dfrac{\sqrt{x+3}-3}{x-6}\)
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2x-6}{4-x}\)
e) \(\lim\limits_{x\rightarrow+\infty}\dfrac{17}{x^2+1}\)
f) \(\lim\limits_{x\rightarrow+\infty}\dfrac{-2x^2+x-1}{3+x}\)
a)
= 
= -4.
b)
= 
= 
(2-x) = 4.
c)
= 
= 
= 
.
=
d)
= 
= -2.
e)
= 0 vì 
(x2 + 1) = 
x2( 1 + 
) = +∞.
f)
= 
\(\lim\limits_{x\rightarrow+\infty}\dfrac{2x-\sqrt{3x^2+2}}{5x+\sqrt{x^2+1}}\)
\(\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{x^2+1}{2x^4+x^2-3}}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}-\sqrt{\dfrac{3x^2}{x^2}+\dfrac{2}{x^2}}}{\dfrac{5x}{x}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}=\dfrac{2-\sqrt{3}}{5+1}=\dfrac{2-\sqrt{3}}{6}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{\dfrac{x^2}{x^4}+\dfrac{1}{x^4}}{\dfrac{2x^4}{x^4}+\dfrac{x^2}{x^4}-\dfrac{3}{x^4}}}=0\)
3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{x^4}{x^6}+\dfrac{1}{x^6}}}{\sqrt{\dfrac{x^4}{x^4}+\dfrac{x^3}{x^4}+\dfrac{1}{x^4}}}=-1\)
Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
a: \(\lim_{x\to+\infty}\frac{\sqrt{x^6+2}}{3x^3-1}=\lim_{x\to+\infty}\frac{x^3\left(\sqrt{1+\frac{2}{x^6}}\right)}{x^3\left(3-\frac{1}{x^3}\right)}\)
\(=\lim_{x\to+\infty}\frac{\sqrt{1+\frac{2}{x^6}}}{3-\frac{1}{x^3}}=\frac13\)
c: \(2x^3-5x-4=2\cdot1^3-5\cdot1-4=2-5-4=2-9=-7<0\)
\(\left(x+1\right)^2=\left(1+1\right)^2=2^2=4\)
\(\lim_{x\to1}\frac{2x^3-5x-4}{\left(x+1\right)^2}=\frac{-7}{4}\)
1) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-1}{x}\)
2)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-x^3+3x-4}{x-1}\)
3) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}\)
4) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}\)
5) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}\)
6) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}\)
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{3x^3+1}-\sqrt{2x^2+x+1}}{\sqrt[4]{4x^4+2}}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2-3x+4}-2x}{\sqrt{x^2+x+1}-x}\)
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow2}\dfrac{x+3}{x^2+x+4}\)
b) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2+5x+6}{x^2+3x}\)
c) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}\)
d) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
e) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+3}{3x-1}\)
f) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-2x+4}-x}{3x-1}\)
\(\lim\limits_{x\rightarrow0^-}\left(\dfrac{1}{x^2}-\dfrac{2}{x^3}\right)\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{1}{x^3-1}-\dfrac{1}{x-1}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt[3]{1-x^3}\right)\)
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
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