Bài 10:
\(\lim u_n=\lim \frac{2^n-5.7^{n+1}}{2^n+7^n}=\lim \frac{(\frac{2}{7})^n-35}{(\frac{2}{7})^n+1}=\frac{0-35}{0+1}=-35\)
(nhớ rằng \(|\frac{2}{7}|< 1\Rightarrow \lim (\frac{2}{7})^n=0\) )
Đáp án C.
Bài 11:
\(\lim u_n=\lim \frac{3^n-2.5^{n+1}}{2^{n+1}+5^n}=\lim \frac{(\frac{3}{5})^n-10}{2(\frac{2}{5})^n+1}=\frac{0-10}{2.0+1}=-10\)
Đáp án A.
Bài 12:
\(\lim u_n=\lim \sqrt[3]{\frac{5-8n}{n+3}}=\lim \sqrt[3]{\frac{\frac{5}{n}-8}{1+\frac{3}{n}}}=\sqrt[3]{-8}=-2\)
Đáp án B.
Bài 13:
\(\lim u_n=\lim \frac{8n+\sin n}{4n+3}\). Trong đó:
\(\lim \frac{8n}{4n+3}=\lim \frac{8}{4+\frac{3}{n}}=\frac{8}{4}=2\)
\(0\leq |\frac{\sin n}{4n+3}|\leq \frac{1}{4n+3}\) mà \(\lim \frac{1}{4n+3}=0\) nên \(\lim \frac{\sin n}{4n+3}=0\)
Do đó: \(\lim u_n=2+0=2\)
Đáp án B.
Bài 14:
\(\lim \frac{1-3n-5n^2}{\cos n+n^2}=\lim \frac{\frac{1}{n^2}-\frac{3}{n}-5}{\frac{\cos n}{n^2}+1}\)
Trong đó: \(0\leq |\frac{\cos n}{n^2}|\leq \frac{1}{n^2}; \lim \frac{1}{n^2}=0\) nên \(\lim \frac{\cos n}{n^2}=0\)
\(\lim \frac{1}{n^2}=0; \lim \frac{3}{n}=0\) (dễ thấy)
Do đó: \(\lim \frac{1-3n-5n^2}{\cos n+n^2}=\frac{-5}{1}=-5\)
Đáp án D.
Bài 16:
\(\frac{S}{3}=\lim [(\frac{-1}{3})^2+(\frac{-1}{3})^3+(\frac{-1}{3})^4+....+(\frac{-1}{3})^{n+1}]] \)
\(\frac{-S}{9}=\lim [(\frac{-1}{3})^3+(\frac{-1}{3})^4+(\frac{-1}{3})^5+...+(\frac{-1}{3})^{n+2}]\)
\(\frac{S}{3}+\frac{S}{9}=\lim [(-\frac{1}{3})^2-(\frac{-1}{3})^{n+2}]\)
\(\frac{4}{9}S=\lim [\frac{1}{9}-(\frac{-1}{3})^{n+2}]\Rightarrow S= \lim [\frac{1}{4}-\frac{9}{4}(-\frac{1}{3})^{n+2}]=\frac{1}{4}\)
Đáp án A.
Bài 15:
\(S=\lim [1+\frac{2}{3}+(\frac{2}{3})^2+(\frac{2}{3})^3+...+(\frac{2}{3})^n]\)
\(S(1-\frac{2}{3})=\lim [(1-\frac{2}{3})[1+\frac{2}{3}+(\frac{2}{3})^2+...+(\frac{2}{3})^n]=\lim [1-(\frac{2}{3})^{n+1}]\)
\(S=\lim [3-3(\frac{2}{3})^{n+1}]=3\)
Đáp án A.
Bài 18:
\(T=1-\tan ^2x+\tan ^4x-\tan ^6x+...+(-1)^n\tan ^{2n}x\)
\(\tan ^2x.T=\tan ^2x-\tan ^4x+\tan ^6x-\tan ^8x+...+(-1)^n\tan ^{2n+2}x\)
\(T(\tan ^2x+1)=1+(-1)^n\tan ^{2n+2}x\Rightarrow T=\frac{1+(-1)^n\tan ^{2n+2}x}{\tan ^2x+1}\)
Với $0< x< \frac{\pi}{4}$ thì $|\tan x|< 1$
$\Rightarrow \lim \tan ^{2n+2}x=0$
Do đó:
$S=\lim T=\lim \frac{1+(-1)^n\tan ^{2n+2}x}{\tan ^2x+1}=\frac{1}{\tan ^2x+1}=\cos ^2x$
Đáp án B.
Bài 19:
\(T=1+\cos ^2x+\cos ^4x+...+\cos ^{2n}x\)
\(\cos ^2x.T=\cos ^2x+\cos ^4x+\cos ^6x+...+\cos ^{2n+2}x\)
\(T(1-\cos ^2x)=1-\cos ^{2n+2}x\)
\(T=\frac{1-\cos ^{2n+2}x}{1-\cos ^2x}=\frac{1-\cos ^{2n+2}x}{\sin ^2x}\)
Với $0< x< \frac{\pi}{2}$ thì $|\cos x|< 1$
$\Rightarrow \lim \cos ^{2n+2}x=0$
$S=\lim T=\lim \frac{1-\cos ^{2n+2}x}{\sin ^2x}=\frac{1}{\sin ^2x}$
Đáp án D.
Bài 20:
$0,(21)=\frac{21}{99}=\frac{7}{33}$. Đáp án A.
Bài 21:
\(0,2(3)=0,2+0,0(3)=\frac{2}{10}+\frac{1}{10}.\frac{3}{9}=\frac{7}{30}\)
Đáp án B.
Bài 22:
\(T=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{2^2}-\frac{1}{3^2})+...+(\frac{1}{2^n}-\frac{1}{3^n})\)
\(=(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})-(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n})\)
\(=\frac{1-(\frac{1}{2})^{n+1}}{1-\frac{1}{2}}-\frac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}}=2[1-(\frac{1}{2})^{n+1}]-\frac{3}{2}[1-(\frac{1}{3})^{n+1}]\)
\(=\frac{1}{2}-2(\frac{1}{2})^{n+1}+\frac{3}{2}(\frac{1}{3})^{n+1}\)
\(\Rightarrow S=\lim T=\frac{1}{2}\)
Đáp án A.
Bài 23:
Ta có:
$|u_n-5|< (\frac{\sqrt{2}}{2})^n< 1$ với mọi $n\in\mathbb{N}$
$\Rightarrow \lim (u_n-5)=0$
$\Rightarrow \lim u_n=5$
Đáp án B.
Bài 24:
\(\lim \frac{v_n}{u_n}=\lim \frac{2(n+1)}{n+2}=\lim \frac{2+\frac{2}{n}}{1+\frac{1}{n}}=\frac{2}{1}=2\)
Đáp án B.
Bài 25:
\(\lim (u_n+v_n)=\lim \frac{(-1)^n+1}{n^2+1}\)
Ta thấy \(|\frac{(-1)^n+1}{n^2+1}|< \frac{2}{n^2+1}; \lim \frac{2}{n^2+1}=0\Rightarrow \lim \frac{(-1)^n+1}{n^2+1}=0\)
Đáp án A.
Bài 26:
Khi $n$ chẵn:
\(\lim u_n=\lim \frac{2n+1}{n+1}=\lim \frac{2+\frac{1}{n}}{1+\frac{1}{n}}=2\)
Khi $n$ lẻ:
\(\lim u_n=\lim \frac{1}{n}=0\)
Do $2\neq 0$ nên $u_n$ không tồn tại giới hạn. Đáp án D.
Bài 17:
Đặt \(T=1-\sin ^2x+\sin ^4x-\sin ^6x+...+(-1)^n\sin ^{2n}x\)
\(\sin ^2x.T=\sin ^2x-\sin ^4x+\sin ^6x-\sin ^8x+...+(-1)^n\sin^ {2(n+1)}x\)
\(T+\sin ^2x.T=1+(-1)^n\sin ^{2n+2}x\)
\(T(\sin ^2x+1)=1+(-1)^n(\sin x)^{2n+2}\Rightarrow T=\frac{1+(-1)^n(\sin x)^{2n+2}}{\sin ^2x+1}\)
Với $0< x< \frac{\pi}{2}$ thì \(|\sin x|< 1\) nên \(\lim (\sin x)^{2n+2}=0\)
Do đó \(S=\lim T=\lim \frac{1+(-1)^n(\sin x)^{2n+2}}{\sin ^2x+1}=\frac{1}{\sin ^2x+1}\)
Đáp án D
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