Tính lim x → 1 x 2 - ( a + 2 ) x + a + 1 x 3 - 1
A. 2 - a 3
B. - 2 - a 3
C. - a 3
D. a 3
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CM
Cao Minh Tâm
27 tháng 9 2019
Đúng(0)
Những câu hỏi liên quan
6 tháng 2 2023
1: \(A=\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
\(=\dfrac{x^2-xa-x+a}{\left(x-a\right)\left(x^2+ax+a^2\right)}\)
\(=\dfrac{\left(x-a\right)\left(x-1\right)}{\left(x-a\right)\left(x^2+ax+a^2\right)}=\dfrac{x-1}{x^2+ax+a^2}\)
\(lim_{x->a}A=lim_{x->a}\left(\dfrac{x-1}{x^2+ax+a^2}\right)\)
\(=\dfrac{a-1}{a^2+a^2+a^2}=\dfrac{a-1}{3a^2}\)
2: \(B=\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\)
\(=\dfrac{-1}{x-1}+\dfrac{3}{x^3-1}\)
\(=\dfrac{-x^2-x-1+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x-2}{x^2+x+1}\)
\(lim_{x->1}\left(B\right)=\dfrac{-1-2}{1^2+1+1}=\dfrac{-3}{3}=-1\)
3: \(C=\dfrac{\left(x+h\right)^3-x^3}{h}=\dfrac{\left(x+h-x\right)\left(x^2+2xh+h^2+x^2+hx+x^2\right)}{h}\)
\(=3x^2+3hx\)
\(lim_{h->0}\left(C\right)=3x^2+3\cdot0\cdot x=3x^2\)
MH
1
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)-\left(a+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-a-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x-a-1}{x^2+x+1}=\dfrac{-a}{3}\)
HT
9 tháng 2 2021
a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)
b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)
AM
9 tháng 2 2022
a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)
b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)
c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)
d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe
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) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
HT
20 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2-x-1}{\sqrt{x^2-x+1}+\sqrt{x^2+x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2x}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}=-\dfrac{2}{1+1}=-1\)
b/ \(=\lim\limits_{x\rightarrow2}\dfrac{4x+1-9}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\dfrac{4}{\left(2+2\right)\left(\sqrt{4.2+1}+3\right)}=\dfrac{1}{6}\)
c/ \(=\lim\limits_{x\rightarrow-2}\dfrac{2x+5-1}{\left(x-2\right)\left(x+2\right)\left(\sqrt{2x+5}+1\right)}=\lim\limits_{x\rightarrow-2}\dfrac{2}{\left(x-2\right)\left(\sqrt{2x+5}+1\right)}=\dfrac{2}{\left(-2-2\right)\left(\sqrt[2]{2.\left(-2\right)+5}+1\right)}=\dfrac{2}{\left(-4\right).2}=-\dfrac{1}{4}\)
ND
1
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2022
\(\lim\limits_{x\rightarrow-3}\dfrac{x^2-1}{x+1}=\dfrac{\left(-3\right)^2-1}{-3+1}=-4\) (câu này \(x\rightarrow-3\) hay \(x\rightarrow-1\) vậy?)
\(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{x+2}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(x+2\right)}{x+2}=\lim\limits_{x\rightarrow-2}\left(2-x\right)=4\)
4 tháng 6 2021
\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)
\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)
\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)
\(=\frac{1^2+1+3}{1}\)
\(=5\)
\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)
\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)
\(=\frac{-1}{2+2}=\frac{-1}{4}\)
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
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