Bài 109206

Question

Tính các giới hạn sau:
a)$\mathop {\lim }\limits_{n \to \infty}\frac{n\sqrt[]{n^3}-3}{5n^2+1} $
b)$\mathop {\lim }\limits_{n \to \infty} \frac{(3n+1)(n+2)(n-1)}{n(2n^2+n+1)} $
c) $\lim\left ( \frac{1}{n+\sqrt[]{1} }+\frac{1}{n+\sqrt[]{2}}+\frac{1}{n+\sqrt[]{3} }+...+\frac{1}{n+\sqrt[]{n} }\right ) $
d)$\mathop {\lim }\limits_{n \to \infty}\frac{1-2^n}{1+2^n} $

score 0 · Answer 1

a) $\infty$

b) $\frac{3}{2} $

c)    $\frac{1}{n+\sqrt[]{1} }+\frac{1}{n+\sqrt[]{2} }+...+\frac{1}{n+\sqrt[]{n}} \leq \frac{1}{n+1}+\frac{1}{n+1}

+...+\frac{1}{n+1}=\frac{n}{n+1} $.
Mặt khác $\frac{1}{n+\sqrt[]{1} }+\frac{1}{n+\sqrt[]{2} }+...+\frac{1}{n+\sqrt[]{n}} \geq \frac{1}{n+\sqrt[]{n} }+\frac

{1}{n+\sqrt[]{n} }+...+\frac{n}{n+\sqrt[]{n} }=\frac{n}{n+\sqrt[]{n} }         $
                     $\frac{n}{n+\sqrt[]{n} } \leq u_n \leq \frac{n}{n+1} \Rightarrow \mathop {\lim }\limits_{n \to \infty}\frac{n}

{n+\sqrt[]{n} } \leq \mathop {\lim }\limits_{n \to \infty}u_n \leq \mathop {\lim }\limits_{n \to \infty}\frac{n}{n+1}$
$\Rightarrow 1 \leq \mathop {\lim }\limits_{n \to \infty}u_n \leq 1$
ĐS: $\mathop {\lim }\limits_{n \to \infty} u_n =1$

d)   $\mathop {\lim }\limits_{n \to \infty}\frac{\frac{1}{2^n}-1 }{\frac{1}{2^n}+1 }=-1 $

Bài 109206

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Bài 109206

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