Cho dãy số U_{n} thỏa mãn\begin{cases}0<U_{1}<1\\ U_{n+1}(1-U_{n})>\frac{1}{4}\end{cases}Tìm Lim U_{n}
Trả lời 25-04-17 07:58 PM
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Cho x(n): \begin{cases}x_{1}=1 \\x_{n+1} =\sqrt{x_{n}(x_{n}+1)(x_{n}+2)(x_{n}+3)+1} \end{cases}Đặt y=\sum_{i=1}^{n}\frac{1}{x_{i}+2}. Tính \mathop {\lim }\limits yn
Trả lời 29-03-17 02:12 AM
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\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[4]{1+4x}-1}{x}
Trả lời 19-02-17 04:30 AM
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\mathop {\lim }\limits_{x \to 1}\frac{2-\sqrt{2x-1}\sqrt[3]{5x+3}}{x-1}
Trả lời 19-02-17 04:02 AM
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tính \mathop {\lim }\limits_{n \to +\infty }\frac{1+4+7+...+(3n-2)}{3n^{2}-4}
Trả lời 19-02-17 03:41 AM
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\mathop {\lim }\limits_{x \to 1} \frac{4x^{6}-5x^{5}+x}{(1-x)^{2}}
Trả lời 16-02-17 08:17 AM
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\mathop {\lim }\limits_{x \to 1} \frac{4x^{6}-5x^{5}+x}{(1-x)^{2}}
Trả lời 16-02-17 07:21 AM
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\mathop {\lim }\limits_{x \to 1} \frac{3x-2-\sqrt{4x^{2}-x-2}}{x^{3}-3x+2}
Trả lời 23-01-17 08:56 PM
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1) \mathop {\lim }\limits_{n \to +\infty }(\sqrt[3]{n^3+n^2}-\sqrt{n^2-1})2) \mathop {\lim }\limits_{n \to +\infty }(\sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3})3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 13-01-17 03:07 AM
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1) \mathop {\lim }\limits_{n \to +\infty }(\sqrt[3]{n^3+n^2}-\sqrt{n^2-1})2) \mathop {\lim }\limits_{n \to +\infty }(\sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3})3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 13-01-17 02:49 AM
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1) \mathop {\lim }\limits_{n \to +\infty }(\sqrt[3]{n^3+n^2}-\sqrt{n^2-1})2) \mathop {\lim }\limits_{n \to +\infty }(\sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3})3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 13-01-17 02:42 AM
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1) \mathop {\lim }\limits_{n \to +\infty }(\sqrt[3]{n^3+n^2}-\sqrt{n^2-1})2) \mathop {\lim }\limits_{n \to +\infty }(\sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3})3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 12-01-17 10:40 PM
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1) \mathop {\lim }\limits_{n \to +\infty }(\sqrt[3]{n^3+n^2}-\sqrt{n^2-1})2) \mathop {\lim }\limits_{n \to +\infty }(\sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3})3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 12-01-17 10:30 PM
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lim \frac{(-1)^n.sin(\frac{\pi n}{6})}{1,000001^n}
Trả lời 03-01-17 08:53 AM
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tính lim\frac{1+3+5+...+(2n-1)}{4n^{2}+6n+1}
Trả lời 24-12-16 05:25 AM
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u_{n}=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{n(n+2)}
Trả lời 24-12-16 04:47 AM
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tìm giới hạn các dãy sau1) \mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt{n^{2}+5n\sqrt{n}}-n}{\sqrt{n-2}}2) \mathop {\lim }\limits_{n \to +\infty }\frac{6n}{n-\sqrt{n^2+4n}}3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 24-12-16 01:42 AM
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tìm giới hạn các dãy sau1) \mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt{n^{2}+5n\sqrt{n}}-n}{\sqrt{n-2}}2) \mathop {\lim }\limits_{n \to +\infty }\frac{6n}{n-\sqrt{n^2+4n}}3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 24-12-16 01:38 AM
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tìm giới hạn các dãy sau1) \mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt{n^{2}+5n\sqrt{n}}-n}{\sqrt{n-2}}2) \mathop {\lim }\limits_{n \to +\infty }\frac{6n}{n-\sqrt{n^2+4n}}3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 24-12-16 01:35 AM
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tìm giới hạn các dãy sau1) \mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt{n^{2}+5n\sqrt{n}}-n}{\sqrt{n-2}}2) \mathop {\lim }\limits_{n \to +\infty }\frac{6n}{n-\sqrt{n^2+4n}}3) $\mathop {\lim }\limits_{n \to +\infty...
Trả lời 24-12-16 01:32 AM
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