Tìm giới hạn D = lim x → 0 ( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x ) x
A. +∞
B. -∞
C. -1/6
D. 6
K
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HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó
\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)
b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)
c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)
LT
1
16 tháng 12 2023
1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)
2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0
3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)
\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)
4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)
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) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
23 tháng 7 2023
a: \(\lim\limits_{x\rightarrow-1^+}x+1=0\)
=>\(\lim\limits_{x\rightarrow-1^+}\dfrac{1}{x+1}=+\infty\)
b: \(\lim\limits_{x\rightarrow-\infty}1-x^2=\lim\limits_{x\rightarrow-\infty}\left[x^2\left(\dfrac{1}{x^2}-1\right)\right]\)
\(=-\infty\)
c: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)
\(\lim\limits_{x\rightarrow3^-}x-3=0\)
\(\lim\limits_{x\rightarrow3^-}-x=3>0\)
=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1
\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)
\(=\dfrac{23}{12}\)
QT
Quoc Tran Anh Le
Giáo viên
22 tháng 9 2023
a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{1 - 3{x^2}}}{{{x^2} + 2x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}\left( {\frac{1}{{{x^2}}} - 3} \right)}}{{{x^2}\left( {1 + \frac{{2x}}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{{{x^2}}} - 3}}{{1 + \frac{2}{x}}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{x^2}}} - \mathop {\lim }\limits_{x \to + \infty } 3}}{{\mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{2}{x}}} = \frac{{0 - 3}}{{1 + 0}} = - 3\)
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{2}{{x + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{2}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}.\mathop {\lim }\limits_{x \to - \infty } \frac{2}{{1 + \frac{1}{x}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}.\frac{{\mathop {\lim }\limits_{x \to - \infty } 2}}{{\mathop {\lim }\limits_{x \to - \infty } 1 + \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}} = 0.\frac{2}{{1 + 0}} = 0\).
PM
1
15 tháng 2 2023
\(=lim_{x->0}\left(\dfrac{1+x^2-1}{x^2\left(\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1\right)}\right)\)
\(=lim_{x->0}1=1\)
AH
Akai Haruma
Giáo viên
5 tháng 11 2023
Lời giải:
1.
\(\lim\limits_{x\to -1}\frac{x^{2019}+1}{x^2+x}=\lim\limits_{x\to -1}\frac{(x+1)(x^{2018}-x^{2017}+x^{2016}-....-x+1)}{x(x+1)}=\lim\limits_{x\to -1}\frac{x^{2018}-x^{2017}+x^{2016}-....-x+1}{x}\)
\(=\frac{(-1)^{2018}-(-1)^{2017}+(-1)^{2016}+....-(-1)+1}{-1}\)
\(=\frac{\underbrace{1+1+....+1+1}_{2019}}{-1}=\frac{2019}{-1}=-2019\)
2.
\(\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+(x^3-1)+....+(x^n-1)}{x-1}\\ =\lim\limits_{x\to 1}\frac{(x-1)+(x-1)(x+1)+(x-1)(x^2+x+1)+....+(x-1)(x^{n-1}+x^{n-2}+...+x+1)}{x-1}\)
$\lim\limits_{x\to 1}[1+(x+1)+(x^2+x+1)+....+(x^{n-1}+x^{n-2}+...+x+1)]$
$=1+2+3+....+n=n(n+1):2$
\(\)
HA
1
Chọn D.
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