Giúp e 3 bài này với ạ 1, Lim sin^2n / n + 2 2, Lim 1 + cosn / 2n + 3 3, Lim cosn + 4 / 5 + n
Chương 4: GIỚI HẠN
tìm giá trị giới hạn \(lim\dfrac{n\sqrt{n}+1}{n^2+2}\)
`lim[n\sqrt{n}+1]/[n^2+2]`
`=lim[n^2\sqrt{1/n}+1]/[n^2+2]`
`=lim[n^2(\sqrt{1/n}+1/[n^2])]/[n^2(1+2/[n^2])]`
`=lim[\sqrt{1/n}+1/[n^2]]/[1+2/[n^2]]`
`=0/1=0`
Đúng 1
Bình luận (0)
Giúp em giải chi tiết 2 câu này với ạ. Em cảm ơn Lim(3√n^3+4 - 3√n^3 -1) căn bậc 3 ạ
\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)
\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)
Đúng 1
Bình luận (1)
\(1.lim\left(\sqrt[3]{8n^3+4n^2+1}-\sqrt[3]{8n^3-2}\right)\)
\(2.lim\left(\sqrt[3]{n^3+n^2+1}+\sqrt[3]{8-n^3}\right)\)
\(3.lim\left(\sqrt[3]{n^3+n^2+2}-n\right)\)
Bài 1. Tìm các giới hạn sau:
a) \(\lim\limits\dfrac{-2n+1}{n}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}\)
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
Đúng 1
Bình luận (0)
\(\lim\dfrac{\sqrt{n}+2}{5n-1}=\lim\dfrac{\dfrac{1}{\sqrt{n}}+\dfrac{2}{n}}{5-\dfrac{1}{n}}=\dfrac{0+0}{5-0}=0\)
Đúng 1
Bình luận (0)
Cho dãy an xác định \(\left\{{}\begin{matrix}a_1=\dfrac{1}{2}\\a_{n+1}=\dfrac{a_n^2}{a_n^2-a_n+1}\end{matrix}\right.\) với n=1,2,...
Tính \(\lim\limits_{n\rightarrow+\infty}a_1+a_2+...+a_n=?\)
giải chi tiết giúp e câu b này ạ ! E cảm ơn
\(lim_{x\rightarrow1}\dfrac{x^2-2x+1}{x^2+3x-4}\) \(=lim_{x\rightarrow1}\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+4\right)}=lim_{x\rightarrow1}\dfrac{x-1}{x+4}=0\)
Đúng 2
Bình luận (0)
\(\lim_(x->1)(\sqrt(10x-9)-1)/(x-1)\)
\(=\lim\limits_{x->1}\dfrac{10x-9-1}{\sqrt{10x-9}+1}\cdot\dfrac{1}{x-1}=\lim\limits_{x->1}\dfrac{10}{\sqrt{10x-9}+1}\)
\(=\dfrac{10}{1+1}=5\)
Đúng 0
Bình luận (0)
Tìm lim
\(\lim\limits_{x\rightarrow\infty}x+1-\sqrt{x^2+3x}\)
Lời giải:
\(\lim\limits_{x\to +\infty}(x+1-\sqrt{x^2+3x})=\lim\limits_{x\to +\infty}\frac{(x+1)^2-(x^2+3x)}{x+1+\sqrt{x^2+3x}}\)
\(=\lim\limits_{x\to +\infty}\frac{-x+1}{x+1+\sqrt{x^2+3x}}=\lim\limits_{x\to +\infty}\frac{-1+\frac{1}{x}}{1+\frac{1}{x}+\sqrt{1+\frac{3}{x}}}=\frac{-1}{1+1}=\frac{-1}{2}\)
\(\lim\limits_{x\to -\infty}(x+1-\sqrt{x^2+3x})=\lim\limits_{x\to -\infty}\frac{(x+1)^2-(x^2+3x)}{x+1+\sqrt{x^2+3x}}\)
\(=\lim\limits_{x\to -\infty}\frac{1-x}{x+1+\sqrt{x^2+3x}}=\lim\limits_{x\to -\infty}\frac{\frac{1}{x}-1}{1+\frac{1}{x}-\sqrt{1+\frac{3}{x}}}\)
\(\lim\limits_{x\to -\infty}(\frac{1}{x}-1)=-1<0\)
\(\lim\limits_{x\to -\infty}(1+\frac{1}{x}-\sqrt{1+\frac{3}{x}})=0\)
\(\Rightarrow \lim\limits_{x\to -\infty}(x+1-\sqrt{x^2+3x})=-\infty\)
Đúng 1
Bình luận (0)