a,b,c thuộc[0;1] tìm giá trị lớn nhất S=a(1-b)+b(1-c)+c(1-a)
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\(=\lim\dfrac{n^2+4n+1-n^2}{\sqrt{n^2+4n+1}+n}=\lim\dfrac{4n+1}{\sqrt{n^2+4n+1}+n}=\lim\dfrac{4+\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+1}=\dfrac{4}{1+1}=2\)
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\(=\lim\dfrac{n^2-\left(n^2+9n-1\right)}{n+\sqrt{n^2+9n-1}}=\lim\dfrac{-9n+1}{n+\sqrt{n^2+9n-1}}=\lim\dfrac{-9+\dfrac{1}{n}}{1+\sqrt{1+\dfrac{9}{n}-\dfrac{1}{n^2}}}=\dfrac{-9}{1+1}=-\dfrac{9}{2}\)
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Do \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}=\dfrac{n^2+n}{2}\)
\(\Rightarrow\lim\dfrac{1+2+...+n}{n^2-1}=\lim\dfrac{n^2+n}{2n^2-2}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{2}{n^2}}=\dfrac{1}{2}\)
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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 \(3^n\))
Chia cả tử và mẫu cho x2, lim đề bài cho sẽ bằng với
\(lim\dfrac{\sqrt{1-\dfrac{1}{n^2}}+3}{\dfrac{1}{n^2}-1}=\dfrac{\sqrt{1-0}+3}{0-1}=-4\)
\(=\lim\dfrac{\sqrt{1-\dfrac{1}{n^2}}+3}{\dfrac{1}{n^2}-1}=\dfrac{1+3}{-1}=-4\)
\(\lim\dfrac{2n^2-1}{4-6n}\) hay \(\lim\dfrac{2n-1}{4-6n}\) vậy nhỉ?
\(a;b;c\in\left[0;1\right]\Rightarrow\left\{{}\begin{matrix}a-1\le0\\b-1\le0\\c-1\le0\end{matrix}\right.\) \(\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\)
\(S=a+b+c-ab-bc-ca=\left(abc-ab-bc-ca+a+b+c-1\right)+1-abc\)
\(S=\left(1-a\right)\left(1-b\right)\left(1-c\right)+1-abc\le1-abc\le1\)
\(S_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\) và các hoán vị