Solve:
$$\mathop {\lim }\limits_{n \to \infty }(\frac{\sqrt[2]{2}+\sqrt[4]{4}+...+\sqrt[2n]{2n}}{1+\sqrt[3]{3}+...+\sqrt[2n-1]{2n-1}})^n$$
ko biết, triển! –  Confusion 21-02-17 05:07 AM
tg bài này có lời giải r ? Hùng Nolan –  ☼SunShine❤️ 20-02-17 02:53 PM
chịu thôi, tại tui lớp 11 –  tran85295 20-02-17 09:44 AM
tui đâu bắt ai phải dùng kiến thức 11 để giải? –  Confusion 20-02-17 09:20 AM
bài này dùng kt lớp 11 là ko thể giải dc –  tran85295 20-02-17 09:16 AM
when????? –  Confusion 20-02-17 07:02 AM
tử < mẫu xong suy ra lim=0 đc k nhỉ –  ๖ۣۜPXM๖ۣۜMinh4212♓ 20-02-17 06:56 AM
+) TH1: $m=n$
Chia $x^n:$
$f(x)=\frac{\frac{a_nx^n}{x^n}+\frac{a_{n-1}x^{n-1}}{x^n}+......+\frac{a_1x}{x^n}+\frac{a_0}{x^n}}{\frac{b_nx^n}{x^n}+........................+\frac{b_0}{x^n}}$
$=> \mathop {\lim }\limits_{x \to +\infty }f(x)=\mathop {\lim }\limits_{x \to +\infty }\frac{an}{bn}=\frac{a}{b}.$
+) TH2: $m>n$
Chia $x^m:$
$f(x)=\frac{\frac{a_nx^n}{x^m}+.................+\frac{a_0}{x^m}}{\frac{b_mx^m}{x^m}+..............\frac{b_0}{x^m}}$
$=>\mathop {\lim }\limits_{x \to +\infty }f(x)=\mathop {\lim }\limits_{x \to +\infty }\frac{0}{b_m}=0$
+) TH3: $m<n$ không có đường tiệm cận $=>\varnothing $
==================================================================
TQ:
$\mathop {\lim }\limits_{x \to \pm \infty }f(x)=\frac{an}{bm}.\mathop {\lim }\limits_{x \to \pm \infty }x^{n-m}$
_____
3TF, mãi mãi một tình yêu <3
điều kiện? –  Confusion 18-04-17 08:43 AM
dk đéo hỉu j –  Lionel Messi 17-04-17 09:38 AM
2 bài khác nhau các bác ạ, chỉ là tiếc :v –  Confusion 17-04-17 01:51 AM

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