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Question

$I_{1}=\mathop {\lim }\limits_{x \to +\infty }\frac{1^{3}+5^{3}+...+(4x-3)^{3}}{(1+5+...+4x-3)^{2}}$

$I_{2}=\mathop {\lim }\limits_{x \to +\infty }\frac{\sin \pi x+4\sqrt[3]{x}}{\sqrt[3]{x}}$

ahihi · Answer 1 · 2/4/2016 9:25:08 PM

Bình chọn tăng 1 Bình chọn giảm

ta có :

$1^3+5^3+...+(4x-3)^3=16x^4-16x^3-2x^2+3x$

$(1+5+...+4x-3)^2=(2x^2-x)^2$

$=> lim=....$

Trả lời 04-02-16 09:25 PM

ahihi
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Thìn em yêu <3 – ๖ۣۜDevilღ 20-02-17 08:52 AM

vắn tắt quá -_- – @_@ *Mèo9119* @_@ 20-02-17 07:54 AM

@_@ *Mèo9119* @_@ · Answer 2 · 2/20/2017 7:54:05 AM

Bình chọn tăng 1 Bình chọn giảm

b)

$I_2=\mathop {\lim }\limits_{x \to +\infty }\frac{\sin \pi x+4\sqrt[3]{x}}{\sqrt[3]{x}}=\mathop {\lim }\limits_{x \to +\infty}(\frac{\sin \pi x}{\sqrt[3]{x}}+4)$

Do

$-1 \leq \sin \pi x \le 1$

=>

$-\frac{1}{\sqrt[3]{x}} \le \frac{\sin \pi x}{\sqrt[3]{x}} \le \frac{1}{\sqrt[3]{x}}$

Mà

$\mathop {\lim }\limits_{x \to +\infty} \frac{-1}{\sqrt[3]{x}}=\mathop {\lim }\limits_{x \to +\infty} \frac{1}{\sqrt[3]{x}}=0$

=>

$\mathop {\lim }\limits_{x \to +\infty} \frac{\sin \pi x}{\sqrt[3]{x}}=0$

=>

$I_2=4$

Trả lời 20-02-17 07:54 AM

@_@ *Mèo9119* @_@
76 2

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@_@ *Mèo9119* @_@ · Answer 3 · 2/20/2017 7:46:25 AM

Bình chọn tăng 2 Bình chọn giảm

a) CM = qui nạp:

$1^3+5^3+...+(4x-3)^3=16x^4-16x^3-2x^2+3x$

$(1)$

+) x=1 => (1) lđ

+) G/sử (1) đúng vs

$x=k \geq 1$

=>

$1^3+5^3+...+(4k-3)^3=16k^4-16k^3-2k^2+3k$

+) Ta CM (1) cũng đúng vs

$x=k+1$ (cái này tự CM đi)

=> (1) đúng vs mọi n thuộc N*

Có:

$(1+5+...+4x-3)^2=(2x^2-x)^2$

=>

$I_1=\mathop {\lim }\limits_{x \to +\infty }\frac{16x^4-16x^3-2x^2+3x}{(2x^2-x)^2}=\mathop {\lim }\limits_{x \to +\infty}\frac{16-\frac{16}{x}-\frac{2}{x^2}+\frac{3}{x^3}}{(2-\frac{1}{x})^2}=\frac{16}{4}=4$

Trả lời 20-02-17 07:46 AM

@_@ *Mèo9119* @_@
76 2

9K 266K