Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(=\lim\limits\frac{1-\frac{1}{n}}{2+\frac{7}{n}}=\frac{1-0}{2+0}=\frac{1}{2}\)
b/ \(=lim\frac{4-\frac{1}{n}+\frac{1}{n^2}}{6+\frac{1}{n^2}}=\frac{4-0+0}{6+0}=\frac{4}{6}=\frac{2}{3}\)
c/ \(=lim\frac{3-\frac{1}{n}}{\frac{1}{n^2}-1}=\frac{3-0}{0-1}=\frac{3}{-1}=-3\)
d/ \(=lim\frac{\frac{8}{n}+\frac{1}{n^2}}{1-\frac{2}{n}+\frac{19}{n^2}}=\frac{0+0}{1-0+0}=\frac{0}{1}=0\)
e/ \(=lim\frac{\sqrt{9-\frac{4}{n^2}}+2}{2+\frac{7}{n}}=\frac{\sqrt{9}+2}{2+0}=\frac{5}{2}\)
Chụp ảnh hoặc sử dụng gõ công thức nhé bạn. Để vầy khó hiểu lắm
\(lim\frac{\sqrt{9n^2+2n}+n-2}{\sqrt{4n^2+1}}=lim\frac{\sqrt{9+\frac{2}{n}}+1-\frac{2}{n}}{\sqrt{4+\frac{1}{n^2}}}=\frac{\sqrt{9}+1}{\sqrt{4}}=2\)
\(lim\frac{n}{\sqrt{4n^2+2}+\sqrt{n^2}}=lim\frac{1}{\sqrt{4+\frac{2}{n^2}}+\sqrt{1}}=\frac{1}{\sqrt{4}+\sqrt{1}}=\frac{1}{3}\)
\(lim\frac{\sqrt{4n+2}-\sqrt{2n-5}}{\sqrt{n+3}}=lim\frac{\sqrt{4+\frac{2}{n}}-\sqrt{2-\frac{5}{n}}}{\sqrt{1+\frac{3}{n}}}=\frac{2-\sqrt{2}}{1}=2-\sqrt{2}\)
l\\(lim\frac{\sqrt{4n^2+n+1}-n}{n^2+2}=lim\frac{\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}-1}{n+\frac{2}{n}}=\frac{1}{\infty}=0\)
\(lim\frac{\sqrt{9n^2+n+1}-2n}{3n^2+2}=\frac{\sqrt{9+\frac{1}{n}+\frac{1}{n^2}}-2}{3n+\frac{2}{n}}=\frac{1}{\infty}=0\)
Muốn giúp bạn lắm mà ko sao dịch được đề :D
Bạn sử dụng công cụ gõ công thức, nó ở ngoài cùng bên trái khung soạn thảo, chỗ khoanh đỏ ấy, cực dễ sử dụng
1.
\(\lim \frac{3n^2+5n+4}{2-n^2}=\lim \frac{\frac{3n^2+5n+4}{n^2}}{\frac{2-n^2}{n^2}}=\lim \frac{3+\frac{5}{n}+\frac{4}{n^2}}{\frac{2}{n^2}-1}=\frac{3}{-1}=-3\)
2.
\(\lim \frac{2n^3-4n^2+3n+7}{n^3-7n+5}=\lim \frac{\frac{2n^3-4n^2+3n+7}{n^3}}{\frac{n^3-7n+5}{n^3}}=\lim \frac{2-\frac{4}{n}+\frac{3}{n^2}+\frac{7}{n^3}}{1-\frac{7}{n^2}+\frac{5}{n^3}}=\frac{2}{1}=2\)
3.
\(\lim (\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1})=\lim (n-\frac{3n}{2n^2+3}+\frac{1}{5}-n-\frac{1}{5n+1})\)
\(=\frac{1}{5}-\lim (\frac{3n}{2n^2+3}+\frac{1}{5n+1})=\frac{1}{5}-\lim (\frac{3}{2n+\frac{3}{n}}+\frac{1}{5n+1})=\frac{1}{5}-0=\frac{1}{5}\)
4.
\(\lim \frac{1+3^n}{4+3^n}=\lim (1-\frac{3}{4+3^n})=1-\lim \frac{3}{4+3^n}=1-0=1\)
5.
\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)
\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)
a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)
\(=lim\left(\frac{n}{\sqrt[3]{\left(n-n^3\right)^2}-n\sqrt[3]{\left(n-n^3\right)}+n^2}+\frac{3n}{\sqrt{n^2+3n}+n}\right)\)
\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)
b/ \(lim\left(\frac{-2\sqrt{n}-4}{\sqrt{n-2\sqrt{n}}+\sqrt{n+4}}\right)=lim\left(\frac{-2-\frac{4}{\sqrt{n}}}{\sqrt{1-\frac{2}{\sqrt{n}}}+\sqrt{1+\frac{4}{n}}}\right)=-\frac{2}{1+1}=-1\)
c/ \(lim\left(\frac{3n^2}{\sqrt[3]{n^6+6n^5+9n^4}+\sqrt[3]{n^6+3n^5}+n^2}\right)=lim\left(\frac{3}{\sqrt[3]{1+\frac{6}{n}+\frac{9}{n^2}}+\sqrt[3]{1+\frac{3}{n}}+1}\right)=\frac{3}{3}=1\)
d/ \(lim\left(\sqrt[3]{n^3+6n}-n+n-\sqrt{n^2-4n}\right)=lim\left(\frac{6n}{\sqrt[3]{n^6+12n^4+36n^2}+\sqrt[3]{n^6+6n^4}+n^2}+\frac{4n}{n+\sqrt{n^2-4n}}\right)\)
\(=lim\left(\frac{6}{\sqrt[3]{n^3+12n+\frac{36}{n}}+\sqrt[3]{n^3+6n}+n}+\frac{4}{1+\sqrt{1-\frac{4}{n}}}\right)=0+\frac{4}{1+1}=2\)
e/ \(lim\left(\frac{-3.3^n+4.4^n}{5.3^n+\frac{3}{2}.4^n}\right)=lim\left(\frac{-3\left(\frac{3}{4}\right)^n+4}{5.\left(\frac{3}{4}\right)^n+\frac{3}{2}}\right)=\frac{0+4}{0+\frac{3}{2}}=\frac{8}{3}\)
f/ \(lim\left(\frac{9^n-5.5^n+7.7^n}{9.3^n+5^n+2.8^n}\right)=lim\left(\frac{1-5.\left(\frac{5}{9}\right)^n+7\left(\frac{7}{9}\right)^n}{9.\left(\frac{1}{3}\right)^n+\left(\frac{5}{9}\right)^n+2.\left(\frac{8}{9}\right)^n}\right)=\frac{1}{0}=+\infty\)
g/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)
\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)
\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)
\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)
\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)