Câu 1.

Vì \(\sqrt{2},\left(\sqrt{2}\right)^2,...,\left(\sqrt{2}\right)^n\) lập thành cấp số nhân có \(u_1=\sqrt{2}=q\) nên

\({u_n} = \sqrt 2 .\dfrac{{1 - {{\left( {\sqrt 2 } \right)}^n}}}{{1 - \sqrt 2 }} = \left( {2 - \sqrt 2 } \right)\left[ {{{\left( {\sqrt 2 } \right)}^n} - 1} \right] \to \lim {u_n} = + \infty \) vì \(\left\{{}\begin{matrix}a=2-\sqrt{2}>0\\q=\sqrt{2}>1\end{matrix}\right.\)

Câu 3.

Ta có biến đổi:

\(\lim \left( {\dfrac{{{n^2} - n}}{{1 - 2{n^2}}} + \dfrac{{2\sin {n^2}}}{{\sqrt n }}} \right) = \lim \dfrac{{{n^2} - n}}{{1 - 2{n^2}}} = \dfrac{1}{2}\)

Câu 4.

\(\lim \dfrac{{1 + 2 + 3 + ... + n}}{{{n^2} + 2}} = \lim \dfrac{{n\left( {n + 1} \right)}}{{2\left( {{n^2} + 2} \right)}} = \dfrac{1}{2}\)

\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)

\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)

\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)

\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)

a/ \(=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}{\dfrac{1}{\sqrt{n}}+\sqrt{\dfrac{n}{n}}}=1\)

b/ \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(\Rightarrow\lim\limits\dfrac{n\left(n+1\right)}{2n^2+4}=\lim\limits\dfrac{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}{\dfrac{2n^2}{n^2}+\dfrac{4}{n^2}}=\dfrac{1}{2}\)

c/ \(=\lim\limits\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{n+1}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{\dfrac{n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}+\dfrac{1}{n^2}}+\dfrac{n}{n}}=\dfrac{1}{1+1}=\dfrac{1}{2}\)

d/ \(=\lim\limits\left[\sqrt{n}\left(\sqrt{3-\dfrac{1}{\sqrt{n}}}-\sqrt{2-\dfrac{1}{\sqrt{n}}}\right)\right]=\lim\limits\left[\sqrt{n}\left(\sqrt{3}-\sqrt{2}\right)\right]=+\infty\)

e/ \(=\lim\limits\dfrac{n^3+2n^2-n-n^3}{\left(\sqrt[3]{n^3+2n^2}\right)^2+n.\sqrt[3]{n^3+2n^2}+n^2}=\lim\limits\dfrac{2n^2-n}{\left(n^3+2n^2\right)^{\dfrac{2}{3}}+n.\left(n^3+2n^2\right)^{\dfrac{1}{3}}+n^2}\)

\(=\dfrac{2}{1+1+1}=\dfrac{2}{3}\)

g/ \(=\lim\limits\dfrac{2^n+9.3^n}{4.3^n+8.2^n}=\lim\limits\dfrac{\left(\dfrac{2}{3}\right)^n+9.\left(\dfrac{3}{3}\right)^n}{4.\left(\dfrac{3}{3}\right)^n+8.\left(\dfrac{2}{3}\right)^n}=\dfrac{9}{4}\)

a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả

b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)

c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)

d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)

e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)

f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)

g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)

\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)

\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)

a) lim \(\left(-3n^3+n^2-1\right)\)

minh le oi ban dao mau so cua ban len cho tu uong roi thay vi tri cua mau thanh n³ +2n

\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)

\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)

\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)

\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)

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