Câu 1:
Giới hạn lim\(\dfrac{5\sqrt{3n^2+n}}{2\left(3n+2\right)}=\dfrac{a\sqrt{3}}{b}\)(a/b) khi đó tổng a+b bằng?
Câu 2:
Cho a và b là các số thực khác 0. Nếu limx->2 \(\dfrac{x^2+ax+b}{x-2}=6\) thì a+b bầng?
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NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
HT
7 tháng 2 2021
\(\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{3x}{x^2}}+\dfrac{ax}{x}}{\dfrac{bx}{x}-\dfrac{1}{x}}=\dfrac{a-1}{b}=3\)
=> A
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
CY
2
MH
12 tháng 2 2022
\(a,lim\left(\sqrt{n^2+n+1}-n\right)\)
\(=lim\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\)
\(=lim\dfrac{1+\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{\sqrt[]{n^3+2n}-2n^2}{3n+1}=\lim\dfrac{\sqrt[]{n+\dfrac{2}{n}}-2n}{3+\dfrac{1}{n}}=\lim\dfrac{n\left(\sqrt[]{\dfrac{1}{n}+\dfrac{2}{n^3}}-2\right)}{3+\dfrac{1}{n}}\)
\(=\dfrac{+\infty\left(0-2\right)}{3}=-\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{n\sqrt{1+2+...+2n}}{3n^2+n-2}=\lim\dfrac{n\sqrt{\dfrac{2n\left(2n+1\right)}{2}}}{3n^2+n-2}=\lim\dfrac{\sqrt{2+\dfrac{1}{n}}}{3+\dfrac{1}{n}-\dfrac{2}{n^2}}=\dfrac{\sqrt{2}}{3}\)
HT
16 tháng 2 2021
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)
\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)
\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)
\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)
AM
11 tháng 2 2022
a. ĐKXĐ: \(n\ge0\)
\(lim_{n\rightarrow0}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=\dfrac{\sqrt{2.0+1}}{\sqrt{8.0}+1}=1\)
\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim_{n\rightarrow+\infty}\dfrac{\sqrt{2+\dfrac{1}{n}}}{\sqrt{8}+\dfrac{1}{\sqrt{n}}}=\dfrac{1}{2}\)
b. ĐKXĐ: \(\left\{{}\begin{matrix}n\ne0\\n\le\dfrac{-1-\sqrt{21}}{2}\\n\ge\dfrac{-1+\sqrt{21}}{2}\end{matrix}\right.\)
\(lim_{n\rightarrow+\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow+\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-2\)
\(lim_{n\rightarrow-\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow-\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-1\)
HP
18 tháng 2 2022
a, \(lim\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim\dfrac{\sqrt{n}.\sqrt{2+\dfrac{1}{n}}}{\sqrt{n}\left(\sqrt{8}+\dfrac{1}{n}\right)}=\dfrac{\sqrt{2}}{\sqrt{8}}=\dfrac{1}{2}\)
1.
\(\lim\dfrac{5\sqrt{3n^2+n}}{2\left(3n+2\right)}=\lim\dfrac{5\sqrt{3+\dfrac{1}{n}}}{2\left(3+\dfrac{2}{n}\right)}=\dfrac{5\sqrt{3}}{6}\Rightarrow a+b=11\)
2.
\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax+b}{x-2}=6\) khi \(x^2+ax+b=0\) có nghiệm \(x=2\)
\(\Rightarrow4+2a+b=0\Rightarrow b=-2a-4\)
\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax-2a-4}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)+a\left(x-2\right)}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+a+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\left(x+a+2\right)=a+4\Rightarrow a+4=6\Rightarrow a=2\Rightarrow b=-8\)
\(\Rightarrow a+b=-6\)