giúp e tự luận với ah
\(\lim\limits_{ }\left(\frac{2}{7}+\frac{16}{49}+...+\frac{5^n-3^n}{7^n}\right)\)
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
12 tháng 3 2020
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
AH
Akai Haruma
Giáo viên
12 tháng 3 2020
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)
\(b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}\)
Ta có:
\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)
\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)
\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)
\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 1:
\(a=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(b=\frac{1-5+1}{0}=\frac{-3}{0}=-\infty\)
\(c=\lim\limits_{x\rightarrow1}\frac{x\left(1+2x\right)\left(1+3x\right)+2x\left(1+3x\right)+3x}{x}=\lim\limits_{x\rightarrow1}\left[\left(1+2x\right)\left(1+3x\right)+2\left(1+3x\right)+3\right]=1+2+3=6\)
\(d=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-1}{5x^4+2x}=\frac{4}{0}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 2:
\(a=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(b=\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
\(c=\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}=\frac{-n}{-1}=n\)
\(\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)=x\left(1+2x\right)...\left(1+nx\right)+\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+\left(1+3x\right)...\left(1+nx\right)\)
\(=...\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx+1\)
\(\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx}{x}\)
\(=\lim\limits_{x\rightarrow0}\left[\left(1+2x\right)...\left(1+nx\right)+2\left(1+3x\right)...\left(1+nx\right)+...+n\right]\)
\(=1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 4 2020
Bài 1:
a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}\)
b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)
c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6\)
d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}\)
\(=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10\)
Bài 2:
\(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2020
a/ Do \(x\rightarrow-3^+\) nên \(x>-3\Rightarrow x+3>0\Rightarrow\left|x+3\right|=x+3\)
\(\Rightarrow\lim\limits_{x\rightarrow-3^+}\frac{3x+9}{\left|x+3\right|}=\lim\limits_{x\rightarrow-3^+}\frac{3\left(x+3\right)}{x+3}=3\)
b/ \(=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(1-3\sqrt{x}\right)}{\sqrt{x}\left(4\sqrt{x}-2\right)}=\lim\limits_{x\rightarrow0^+}\frac{1-3\sqrt{x}}{4\sqrt{x}-2}=-\frac{1}{2}\)
Ở câu này \(x\rightarrow0^+\) có nghĩa \(x>0\), nó chỉ để căn thức xác định, ngoài ra ko có gì đặc biệt hết
c/ Tương tự câu c, cũng chỉ để căn thức xác định \(\left(x< 1\right)\)
\(\lim\limits_{x\rightarrow1^-}\frac{\sqrt{1-x}}{\left(1-x\right)\left(x+4\right)}=\lim\limits_{x\rightarrow1^-}\frac{1}{\sqrt{1-x}\left(x+4\right)}=+\infty\)
d/ Chắc bạn ghi nhầm đề, đây ko phải giới hạn dạng vô định (vì tử khác 0, mẫu bằng 0):
\(x\rightarrow\sqrt{2}^-\Rightarrow x< \sqrt{2}\Rightarrow x^4-4< 0\)
\(\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}^-}\frac{\left|x-2\right|}{x^4-4}=-\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2020
\(a=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{9+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{9+\sqrt{x+7}}}{x-1}=\frac{29}{36}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(2-\frac{3}{x}\right)^2.x^3\left(4+\frac{7}{x}\right)^3}{x^3\left(3+\frac{1}{x^3}\right).x^2\left(10+\frac{9}{x^2}\right)}=\frac{2.4}{3.10}=\frac{4}{15}\)
\(c=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\frac{\frac{-4x^2}{\sqrt{1+4x}+2x+1}+\frac{8x^3+12x^2}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{-4}{\sqrt{1+4x}+2x+1}+\frac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=\frac{-4}{1+1}+\frac{12}{1+1+1}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2020
\(d=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}\left(\sqrt{1+4x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{4x\sqrt{1+6x}}{x\left(\sqrt{1+4x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{6x}{x\left(\sqrt{1+6x}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\frac{4\sqrt{1+6x}}{\sqrt{1+4x}+1}+\lim\limits_{x\rightarrow0}\frac{6}{\sqrt{1+6x}+1}=\frac{4}{1+1}+\frac{6}{1+1}=5\)
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}\left(\sqrt{1+2x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{2x\sqrt[3]{1+4x}}{x\left(\sqrt{1+2x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{4x}{x\left(\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\frac{2\sqrt[3]{1+4x}}{\sqrt{1+2x}+1}+\lim\limits_{x\rightarrow0}\frac{4}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}=\frac{2}{1+1}+\frac{4}{1+1+1}=\frac{7}{3}\)
8 tháng 8 2022
1) Có \(u_{n+1}-u_n=\dfrac{1}{2}u^2_n-2u_n+2=\dfrac{1}{2}\left(u_n-2\right)^2\) (1)
+) CM \(u_n>2\) (n thuộc N*)
n=1 : u1= 5/2 > 2 (đúng)
Giả sử n=k, uk > 2 (k thuộc N*)
Ta cần CM n = k + 1. Thật vậy ta có:
\(u_{k+1}=\dfrac{1}{2}u^2_k-u_k+2=\dfrac{1}{2}\left(u_k-2\right)^2+u_k\) (đúng)
Vậy un > 2 (n thuộc N*) (2)
Từ (1) (2) => un+1 - un > 0, hay un+1 > un
=> (un) là dãy tăng => \(\lim\limits_{n\rightarrow\infty}u_n=+\infty\)
2) \(2u_{n+1}=u^2_n-2u_n+4\)
\(\Leftrightarrow2u_{n+1}-4=u^2_n-2u_n\)
\(\Leftrightarrow2\left(u_{n+1}-2\right)=u_n\left(u_n-2\right)\)
\(\Leftrightarrow\dfrac{1}{u_{n+1}-2}=\dfrac{2}{u_n\left(u_n-2\right)}=\dfrac{1}{u_n-2}-\dfrac{1}{u_n}\)
\(\Leftrightarrow\dfrac{1}{u_n}=\dfrac{1}{u_n-2}-\dfrac{1}{u_{n+1}-2}\)
\(S=\dfrac{1}{u_1}+\dfrac{1}{u_2}+...+\dfrac{1}{u_n}\)
\(=\dfrac{1}{u_1-2}-\dfrac{1}{u_2-2}+\dfrac{1}{u_2-2}+...-\dfrac{1}{u_{n+1}-2}\)
\(=\dfrac{1}{u_1-2}-\dfrac{1}{u_{n+1}-2}\)
\(=2-\dfrac{1}{u_{n+1}-2}\)
\(\Leftrightarrow\lim\limits_{n\rightarrow\infty}S=2\)
Đặt \(P=\frac{5-3}{7}+\frac{5^2-3^2}{7^2}+...+\frac{5^n-3^n}{7^n}=\frac{5}{7}+\left(\frac{5}{7}\right)^2+...+\left(\frac{5}{7}\right)^n-\frac{3}{7}-\left(\frac{3}{7}\right)^2-...-\left(\frac{3}{7}\right)^n=A-B\)
\(A=\frac{5}{7}+\left(\frac{5}{7}\right)^2+...+\left(\frac{5}{7}\right)^n\) là tổng CSN với \(u_1=\frac{5}{7};q=\frac{5}{7};n=n\)
\(\Rightarrow A=\frac{5}{7}.\frac{1-\left(\frac{5}{7}\right)^{n+1}}{1-\frac{5}{7}}=\frac{5}{2}-\frac{5}{2}.\left(\frac{5}{7}\right)^{n+1}\)
\(B=\frac{3}{7}+\left(\frac{3}{7}\right)^2+...+\left(\frac{3}{7}\right)^n\) là tổng CSN với \(u_1=\frac{3}{7};q=\frac{3}{7}\)
\(\Rightarrow B=\frac{3}{7}.\frac{1-\left(\frac{3}{7}\right)^{n+1}}{1-\frac{3}{7}}=\frac{3}{4}-\frac{3}{4}.\left(\frac{3}{7}\right)^{n+1}\)
\(\Rightarrow limP=lim\left(\frac{5}{2}-\frac{5}{2}\left(\frac{5}{7}\right)^{n+1}-\frac{3}{4}+\frac{3}{4}\left(\frac{3}{7}\right)^{n+1}\right)=\frac{5}{2}-\frac{3}{4}=\frac{7}{4}\)