giới hạn \(lim\sqrt{\frac{3x^4+4x^5+2}{9x^5+5x^4+4}}\left(x\rightarrow+\infty\right)\) bằng :
A. 0
B. 2/3
C. \(\sqrt{\frac{5}{3}}\)
D. \(\sqrt{\frac{1}{3}}\)
K
Khách
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NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 5 2020
\(L_1=\lim\limits_{x\rightarrow0}\frac{x\left(x^2+3x-2\right)}{x\left(x^4+4\right)}=\lim\limits_{x\rightarrow0}\frac{x^2+3x-2}{x^4+4}=-\frac{1}{2}\)
\(L_2=\lim\limits_{x\rightarrow+\infty}\frac{1-\frac{3}{x^2}+\frac{2}{x^3}}{\left(\frac{4}{x}-2\right)^3}=\frac{1}{\left(-2\right)^3}=-\frac{1}{8}\)
\(L_3=\lim\limits_{x\rightarrow-1}\frac{\left(2x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{2x+1}{x}=1\)
\(L_4=\lim\limits_{x\rightarrow2}\frac{x^2-4x+1}{4-x^2}=\frac{1}{0}=+\infty\)
\(L_5=\lim\limits_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-2}=\frac{0}{1}=0\)
\(L_6=\lim\limits_{x\rightarrow1}\frac{x+3-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{-\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{-\left(x+2\right)}{\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\frac{-3}{2.4}=-\frac{3}{8}\)
\(L_7=\lim\limits_{x\rightarrow+\infty}\frac{x^2+x+1-\left(x-1\right)^2}{\sqrt{x^2+x+1}+x-1}\lim\limits_{x\rightarrow+\infty}\frac{3x}{\sqrt{x^2+x+1}+x-1}=\lim\limits_{x\rightarrow+\infty}\frac{3}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1-\frac{1}{x}}=\frac{3}{2}\)
\(L_8=\lim\limits_{x\rightarrow-\infty}\frac{x^2+x+1-\left(3x-2\right)^2}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8x^2+13x-3}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8+\frac{13}{x}-\frac{3}{x^2}}{-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+3-\frac{2}{x}}=\frac{-8}{-1+3}=-4\)
AH
Akai Haruma
Giáo viên
12 tháng 3 2020
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
AH
Akai Haruma
Giáo viên
12 tháng 3 2020
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
AH
Akai Haruma
Giáo viên
3 tháng 4 2020
Bài 1:
\(\lim\limits _{x\to 1}\frac{4x^6-5x^5+x}{(1-x)^2}=\lim\limits _{x\to 1}\frac{x(x-1)^2(4x^3+3x^2+2x+1)}{(1-x)^2}\)
\(=\lim\limits _{x\to 1}x(4x^3+3x^2+2x+1)=1(4.1^3+3.1^2+2.1+1)=10\)
AH
Akai Haruma
Giáo viên
3 tháng 4 2020
Bài 3:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-(ax+b)]=0\)
\(\Rightarrow \lim\limits _{x\to +\infty}\frac{\sqrt{9x^2-4x+3}-(ax+b)}{x}=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\left(\sqrt{9-\frac{4}{x}+\frac{3}{x^2}}-a+\frac{b}{x}\right)=0\)
\(\Leftrightarrow a=3\)
Thay $a=3$ vào đk ban đầu:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-3x-b]=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty} (\sqrt{9x^2-4x+3}-3x)=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4x+3}{\sqrt{9x^2-4x+3}+3x}=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4+\frac{3}{x}}{\sqrt{9-\frac{4}{x}+\frac{3}{x}}+3}=b\)
\(\Leftrightarrow \frac{-4}{6}=b\Leftrightarrow b=-\frac{2}{3}\)
30 tháng 1 2023
a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)
\(=\dfrac{2x-5}{7}\)
\(=\dfrac{2}{7}x-\dfrac{5}{7}\)
\(=-\infty\)
b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)
\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)
HT
9 tháng 2 2021
Hic nan qua :( Lam vay
P/s: Anh Lam check all ho em nhung bai em lam nhe :( Em cam on
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}=\dfrac{-1}{1+1}=-\dfrac{1}{2}\)
2/ \(=\lim\limits_{x\rightarrow-\infty}x\left(\dfrac{4x^2+1-x^2}{\sqrt{4x^2+1}+x}\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}}{-\sqrt{\dfrac{4x^2}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}=\dfrac{1}{-2+1}=-1\)
3/ \(=\lim\limits_{x\rightarrow-\infty}x^5\left(4-\dfrac{3}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^5}\right)=-\infty\)
4/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{x^4}\left(\sqrt{1-\dfrac{x^3}{x^4}+\dfrac{x^2}{x^4}-\dfrac{x}{x^4}}\right)=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 5 2020
\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x^2+1}+x}{3x+5}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{x^2}}+1}{3+\frac{5}{x}}=\frac{2}{3}\)
Lời giải:
\(\lim\limits _{x\to +\infty}\sqrt{\frac{3x^4+4x^5+2}{9x^5+5x^4+4}}=\lim\limits _{x\to +\infty}\sqrt{\frac{\frac{3}{x}+4+\frac{2}{x^5}}{9+\frac{5}{x}+\frac{4}{x^5}}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
Đáp án B.