Cho các kết quả tính giới hạn sau:
Hỏi có bao nhiêu kết quả đúng trong các kết quả trên?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-8-\dfrac{1}{n^2}\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-3\right)}=\lim\limits_{n\rightarrow\infty}-\dfrac{8}{1-3}\cdot n=\lim\limits_{n\rightarrow\infty}4n=+\infty\)
2:
\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n^2}\right)}{n\left(\sqrt{4+\dfrac{5}{n^2}}-1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\left(\dfrac{3}{\sqrt{4}-1}\right)=+\infty\)
1: \(I=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2-n^2+1}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3}{n\left(\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1-\dfrac{1}{n^2}}\right)}\)
=0
2: \(\lim\limits_{n\rightarrow\infty}\sqrt{n^2+2n+2}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2n+2-n^2}{\sqrt{n^2+2n+2}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n+2}{\sqrt{n^2+2n+2}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{2}{n^2}}-1}\)
\(=+\infty\)
I.
Do \(\left(u_n\right)\) là cấp số nhân \(\Rightarrow\)\(u_4=u_3.q\Rightarrow q=\dfrac{u_4}{u_3}=\dfrac{10}{3}\)
\(u_3=u_1q^2\Rightarrow u_1=\dfrac{u_3}{q^2}=\dfrac{27}{100}\)
2. Công thức số hạng tổng quát: \(u_n=\dfrac{27}{100}.\left(\dfrac{10}{3}\right)^{n-1}\)
II.
1. \(\lim\limits\dfrac{-3n^2+2n-2022}{3n^2-2022}=\lim\dfrac{-3+\dfrac{2}{n}-\dfrac{2022}{n^2}}{3-\dfrac{2022}{n^2}}=\dfrac{-3+0-0}{3-0}=-1\)
2.
\(\lim\limits_{x\rightarrow2}\dfrac{x^2-5x+6}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x-3\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(x-3\right)=-1\)
\(I=\lim\limits\dfrac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}\)
Xet tren tu la 1 csc voi : \(\left\{{}\begin{matrix}u_1=1\\q=a\end{matrix}\right.\Rightarrow S_a=1.\dfrac{a^{n+1}-1}{a-1}\)
Tuong tu cho mau so: \(S_b=1.\dfrac{b^{n+1}-1}{b-1}\)
\(\Rightarrow.....=\lim\limits\dfrac{\dfrac{a^{n+1}-1}{a-1}}{\dfrac{b^{n+1}-1}{b-1}}=\dfrac{\dfrac{1}{a-1}}{\dfrac{1}{b-1}}=\dfrac{1-b}{1-a}\)
Đáp án B
Lý thuyết SGK.