1: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)

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

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+1}\)

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

2: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}\)

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

Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.

\(\lim\left(1+\dfrac{-1}{2^n}\right)=1+0=1\Rightarrow a=1\)

\(\lim\left(\dfrac{n^5}{n^4-2n^3+1}-n\right)=\lim\left(\dfrac{n^5-n\left(n^4-2n^3+1\right)}{n^4-2n^3+1}\right)\)

\(=\lim\left(\dfrac{2n^4-n}{n^4-2n^3+1}\right)=\lim\left(\dfrac{2-\dfrac{1}{n^3}}{1-\dfrac{2}{n}+\dfrac{1}{n^4}}\right)=2\)

3:

\(\lim\limits_{n\rightarrow\infty}\dfrac{2-5^{n-2}}{3^n+2\cdot5^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{5^{n-2}}{5^n}}{\dfrac{3^n}{5^n}+2\cdot\dfrac{5^n}{5^n}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{1}{25}}{\left(\dfrac{3}{5}\right)^n+2\cdot1}\)

\(=-\dfrac{1}{25}:2=-\dfrac{1}{50}\)

1:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)

\(=-\dfrac{4}{1}=-4\)

2:

\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n+1}{2^n-1}\)

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

\(\lim\dfrac{3+4^n}{1+3.4^{n+1}}=\lim\dfrac{3+4^n}{1+12.4^n}=\lim\dfrac{3\left(\dfrac{1}{4}\right)^n+1}{\left(\dfrac{1}{4}\right)^n+12}=\dfrac{0+1}{0+12}=\dfrac{1}{12}\)

\(\lim\dfrac{\left(-2\right)^n+3^n}{\left(-2\right)^{n+1}+3^{n+1}}=\lim\dfrac{\left(-2\right)^n+3^n}{-2\left(-2\right)^n+3.3^n}=\lim\dfrac{\left(-\dfrac{2}{3}\right)^n+1}{-2\left(-\dfrac{2}{3}\right)^n+3}=\dfrac{0+1}{0+3}=\dfrac{1}{3}\)

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

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

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

\(a=\lim\dfrac{\left(n-2\right)!\left(n-1+\left(n-1\right)n\right)}{\left(n-2\right)!\left(\left(n+2\right)\left(n+1\right)n\left(n-1\right)-1\right)}+\lim\dfrac{3}{\left(n+2\right)!-\left(n-2\right)!}\)

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

\(=0+0=0\)

\(b=\lim\dfrac{2+\dfrac{1}{n}}{3^n}=\dfrac{2}{\infty}=0\)