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

Câu hỏi này được treo giải thưởng trị giá +5000 vỏ sò bởi @_@ *Mèo9119* @_@, đã hết hạn vào lúc 03-02-16 10:15 PM

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$
$=> lim=....$
Thìn em yêu <3 –  ๖ۣۜDevilღ 20-02-17 08:52 AM
vắn tắt quá -_- –  @_@ *Mèo9119* @_@ 20-02-17 07:54 AM
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: $(1+5+...+4x-3)^2=(2x^2-x)^2$
=> $I_1=\mathop {\lim }\limits_{x \to +\infty }\frac{16x^4-16x^3-2x^2+3x}{(2x^2-x)^2}=\mathop {\lim }\limits_{x \to +\infty}\frac{16-\frac{16}{x}-\frac{2}{x^2}+\frac{3}{x^3}}{(2-\frac{1}{x})^2}=\frac{16}{4}=4$
vậy ý cậu là ntn? –  Confusion 20-02-17 09:19 AM
ko phải ai cũng biết sai phân :v –  tran85295 20-02-17 09:18 AM
sai phân 🙌 –  Confusion 20-02-17 09:15 AM
cách này có vẻ ko tự nhiên, vì xác định (1) ko phải là luôn làm dc –  tran85295 20-02-17 09:12 AM
hack nick??? @@ –  @_@ *Mèo9119* @_@ 20-02-17 08:13 AM
chị bị hack nick à? @_@ –  Confusion 20-02-17 08:07 AM
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 $\mathop {\lim }\limits_{x \to +\infty} \frac{-1}{\sqrt[3]{x}}=\mathop {\lim }\limits_{x \to +\infty} \frac{1}{\sqrt[3]{x}}=0$
=> $\mathop {\lim }\limits_{x \to +\infty} \frac{\sin \pi x}{\sqrt[3]{x}}=0$
=> $I_2=4$

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