CMR \(lim\dfrac{n^2}{2^n}\)=0
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NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2021
a. Chắc đề là: \(\lim\dfrac{2-5^{n-2}}{3^n+2.5^n}=\lim\dfrac{2\left(\dfrac{1}{5}\right)^{n-2}-1}{9\left(\dfrac{3}{5}\right)^{n-2}+50}=-\dfrac{1}{50}\)
b. \(=\lim\dfrac{2\left(\dfrac{1}{5}\right)^n-25}{\left(\dfrac{3}{5}\right)^n-2}=\dfrac{25}{2}\)
2.
Đặt \(f\left(x\right)=x^4+x^3-3x^2+x+1\)
Hàm f(x) liên tục trên R
\(f\left(0\right)=1>0\) ; \(f\left(-1\right)=-3< 0\)
\(\Rightarrow f\left(0\right).f\left(-1\right)< 0\Rightarrow f\left(x\right)=0\) luôn có ít nhất 1 nghiệm thuộc khoảng \(\left(-1;0\right)\)
Hay pt đã cho luôn có ít nhất 1 nghiệm âm lớn hơn -1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2021
3.
Ta có: M là trung điểm AD, N là trung điểm SD
\(\Rightarrow\) MN là đường trung bình tam giác SAD
\(\Rightarrow MN||SA\Rightarrow\left(MN,SC\right)=\left(SA,SC\right)\)
Ta có: \(AC=\sqrt{AB^2+BC^2}=a\sqrt{2}\)
\(SA=SC=a\)
\(\Rightarrow SA^2+SC^2=AC^2\Rightarrow\Delta SAC\) vuông tại S hay \(SA\perp SC\)
\(\Rightarrow\) Góc giữa MN và SC bằng 90 độ
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)
\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)
\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)
\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)
HT
1 tháng 2 2021
a/ \(=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}{\dfrac{1}{\sqrt{n}}+\sqrt{\dfrac{n}{n}}}=1\)
b/ \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\lim\limits\dfrac{n\left(n+1\right)}{2n^2+4}=\lim\limits\dfrac{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}{\dfrac{2n^2}{n^2}+\dfrac{4}{n^2}}=\dfrac{1}{2}\)
c/ \(=\lim\limits\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{n+1}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{\dfrac{n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}+\dfrac{1}{n^2}}+\dfrac{n}{n}}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
d/ \(=\lim\limits\left[\sqrt{n}\left(\sqrt{3-\dfrac{1}{\sqrt{n}}}-\sqrt{2-\dfrac{1}{\sqrt{n}}}\right)\right]=\lim\limits\left[\sqrt{n}\left(\sqrt{3}-\sqrt{2}\right)\right]=+\infty\)
e/ \(=\lim\limits\dfrac{n^3+2n^2-n-n^3}{\left(\sqrt[3]{n^3+2n^2}\right)^2+n.\sqrt[3]{n^3+2n^2}+n^2}=\lim\limits\dfrac{2n^2-n}{\left(n^3+2n^2\right)^{\dfrac{2}{3}}+n.\left(n^3+2n^2\right)^{\dfrac{1}{3}}+n^2}\)
\(=\dfrac{2}{1+1+1}=\dfrac{2}{3}\)
g/ \(=\lim\limits\dfrac{2^n+9.3^n}{4.3^n+8.2^n}=\lim\limits\dfrac{\left(\dfrac{2}{3}\right)^n+9.\left(\dfrac{3}{3}\right)^n}{4.\left(\dfrac{3}{3}\right)^n+8.\left(\dfrac{2}{3}\right)^n}=\dfrac{9}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2021
\(a=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1+\dfrac{2}{n}}=\dfrac{0}{1}=0\)
\(b=\lim n^3\left(-2+\dfrac{1}{n}+\dfrac{2}{n^3}\right)=+\infty.\left(-2\right)=-\infty\)
\(c=\lim\dfrac{\sqrt{9-\dfrac{1}{n}-\dfrac{1}{n^2}}}{4-\dfrac{2}{n}}=\dfrac{\sqrt{9}}{4}=\dfrac{3}{4}\)
\(d=\lim\dfrac{\left(\dfrac{3}{4}\right)^n+5}{1+\left(\dfrac{2}{4}\right)^n}=\dfrac{5}{1}=5\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 1 2022
1/...
2/ \(=\lim\dfrac{\dfrac{1}{n\sqrt{n}}-1}{4+\dfrac{1}{n^2\sqrt{n}}}=\dfrac{0-1}{4+0}=-\dfrac{1}{4}\) (chia cả tử-mẫu cho \(n^3\))
3/ \(=\lim\dfrac{3-\left(\dfrac{1}{4}\right)^n}{2.\left(\dfrac{3}{4}\right)^n+4\left(\dfrac{1}{4}\right)^n}=\dfrac{3-0}{2.0+3.0}=\dfrac{3}{0}=+\infty\) (chia tử mẫu cho \(4^n\))
4/ \(=\lim\dfrac{2.2^n+\dfrac{4}{3}.3^n}{1-\dfrac{1}{2}.2^n+3.3^n}=\lim\dfrac{2.\left(\dfrac{2}{3}\right)^n+\dfrac{4}{3}}{\left(\dfrac{1}{3}\right)^n-\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^n+3}=\dfrac{2.0+\dfrac{4}{3}}{0-\dfrac{1}{2}.0+3}=\dfrac{4}{9}\) (chia tử mẫu cho \(3^n\))
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
JE
1
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
\(\lim\sqrt{\dfrac{3^n+2.2^n}{sin^n50^0+9.3^n}}=\lim\sqrt{\dfrac{1+2\left(\dfrac{2}{3}\right)^n}{\left(\dfrac{sin50^0}{3}\right)^n+9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
\(2^n=\left(1+1\right)^2=1+C_n^1+C_n^2+C_n^3+...+C_n^n>C_n^3\) (khi n đủ lớn)
\(\Rightarrow2^n>\dfrac{n\left(n-1\right)\left(n-2\right)}{6}\)
\(\Rightarrow\dfrac{n^2}{2^n}< \dfrac{6n^2}{n\left(n-1\right)\left(n-2\right)}=\dfrac{6n}{\left(n-1\right)\left(n-2\right)}\)
Đồng thời do \(\left\{{}\begin{matrix}n^2>0\\2^n>0\end{matrix}\right.\) \(\Rightarrow\dfrac{n^2}{2^n}>0\)
\(\Rightarrow0< \dfrac{n^2}{2^n}< \dfrac{6n}{\left(n-1\right)\left(n-2\right)}\)
Mà \(\lim\left(0\right)=\lim\left(\dfrac{6n}{\left(n-1\right)\left(n-2\right)}\right)=0\)
\(\Rightarrow\lim\left(\dfrac{n^2}{2^n}\right)=0\)