Tìm các giới hạn sau:
\(a,lim\dfrac{3+4^n}{1+3.4^{n+1}}\)
\(b,lim\dfrac{\left(-2\right)^n+3^n}{\left(-2\right)^{n+1}+3^{n+1}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)
\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)
\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)
\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
MH
11 tháng 2 2022
\(a,lim\dfrac{^3\sqrt{8n^3+2n}}{-n+3}\)
\(=lim\dfrac{^3\sqrt{8+\dfrac{2}{n^2}}}{-1+\dfrac{3}{n}}=\dfrac{^3\sqrt{8}}{-1}=\dfrac{2}{-1}=-2\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{\left(2n\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n-1\right)\left(3-2n\right)}=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1-\dfrac{1}{n}\right)\left(\dfrac{3}{n}-2\right)}=\dfrac{2.1}{1.\left(-2\right)}=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 2 2022
\(\lim\dfrac{\left(-3\right)^n-4.5^{n+1}}{2.4^n+3.5^n}=\lim\dfrac{\left(-3\right)^n+20.5^n}{2.4^n+3.5^n}=\lim\dfrac{\left(-\dfrac{3}{5}\right)^n+20}{2\left(\dfrac{4}{5}\right)^n+3}=\dfrac{0+20}{0+3}=\dfrac{20}{3}\)
\(\lim\dfrac{2^n-3^n+4.5^{n+2}}{2^{n+1}+3^{n+2}+5^{n+1}}=\lim\dfrac{2^n-3^n+100.5^n}{2.2^n+9.3^n+5.5^n}=\lim\dfrac{\left(\dfrac{2}{5}\right)^n-\left(\dfrac{3}{5}\right)^n+100}{2\left(\dfrac{2}{5}\right)^n+9\left(\dfrac{3}{5}\right)^n+5}=\dfrac{100}{5}=20\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{\sqrt{n^2+n-1}-n}{2n+3}=\lim\dfrac{n-1}{\left(2n+3\right)\left(\sqrt{n^2+n-1}+n\right)}\)
\(=\lim\dfrac{1-\dfrac{1}{n}}{\left(2+\dfrac{3}{n}\right)\left(\sqrt{n^2+n-1}+n\right)}=\dfrac{1}{2.+\infty}=0\)
AM
12 tháng 2 2022
a. ĐKXĐ: \(n\ne\dfrac{-3}{2}\); \(\left[{}\begin{matrix}x< \dfrac{-1-\sqrt{5}}{2}\\x>\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{n^2+n-1}-n}{2n+3}=\)\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{n}-\dfrac{1}{n^2}}-1}{2+\dfrac{3}{n}}=0\)
\(\lim\dfrac{3+4^n}{1+3.4^{n+1}}=\lim\dfrac{3+4^n}{1+12.4^n}=\lim\dfrac{3\left(\dfrac{1}{4}\right)^n+1}{\left(\dfrac{1}{4}\right)^n+12}=\dfrac{0+1}{0+12}=\dfrac{1}{12}\)
\(\lim\dfrac{\left(-2\right)^n+3^n}{\left(-2\right)^{n+1}+3^{n+1}}=\lim\dfrac{\left(-2\right)^n+3^n}{-2\left(-2\right)^n+3.3^n}=\lim\dfrac{\left(-\dfrac{2}{3}\right)^n+1}{-2\left(-\dfrac{2}{3}\right)^n+3}=\dfrac{0+1}{0+3}=\dfrac{1}{3}\)