5.
\(\lim\dfrac{n+1}{n^2-2}=\lim\dfrac{n^2\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{n^2\left(1-\dfrac{2}{n^2}\right)}=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1-\dfrac{2}{n^2}}=\dfrac{0+0}{1-0}=0\)
\(\lim\dfrac{n\left(n+1\right)}{\left(n+4\right)^3}=\lim\dfrac{n^3\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{n^3\left(1+\dfrac{4}{n}\right)^3}=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{\left(1+\dfrac{4}{n}\right)^3}=\dfrac{0+0}{\left(1+0\right)^3}=0\)
\(\lim\dfrac{3n^3-2n+5}{2n^2+5n-3}=\lim\dfrac{n^3\left(3-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}{n^3\left(\dfrac{2}{n}+\dfrac{5}{n^2}-\dfrac{3}{n^3}\right)}=\lim\dfrac{3-\dfrac{2}{n^2}+\dfrac{5}{n^3}}{\dfrac{2}{n}+\dfrac{5}{n^2}-\dfrac{3}{n^3}}=\dfrac{3}{0}=+\infty\)
\(\lim\dfrac{2n^3}{n^4+3n^2+1}=\lim\dfrac{n^4\left(\dfrac{2}{n}\right)}{n^4\left(1+\dfrac{3}{n^2}+\dfrac{1}{n^4}\right)}=\lim\dfrac{\dfrac{2}{n}}{1+\dfrac{3}{n^2}+\dfrac{1}{n^4}}=\dfrac{0}{1}=0\)
6.
\(\lim\dfrac{3^n-4^n+5^n}{3^n+4^n-5^n}=\lim\dfrac{5^n\left[\left(\dfrac{3}{5}\right)^n-\left(\dfrac{4}{5}\right)^n+1\right]}{5^n\left[\left(\dfrac{3}{5}\right)^n+\left(\dfrac{4}{5}\right)^n-1\right]}=\lim\dfrac{\left(\dfrac{3}{5}\right)^n-\left(\dfrac{4}{5}\right)^n+1}{\left(\dfrac{3}{5}\right)^n+\left(\dfrac{4}{5}\right)^n-1}=\dfrac{0+0+1}{0+0-1}=-1\)
\(\lim\dfrac{1+3^n}{4+3^n}=\lim\dfrac{3^n\left[\left(\dfrac{1}{3}\right)^n+1\right]}{3^n\left[4.\left(\dfrac{1}{3}\right)^n+1\right]}=\lim\dfrac{\left(\dfrac{1}{3}\right)^n+1}{4.\left(\dfrac{1}{3}\right)^n+1}=\dfrac{0+1}{4.0+1}=1\)
\(\lim\dfrac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim\dfrac{7^n\left[4.\left(\dfrac{3}{7}\right)^n+7\right]}{7^n\left[\left(\dfrac{5}{7}\right)^n+1\right]}=\lim\dfrac{4.\left(\dfrac{3}{7}\right)^n+7}{\left(\dfrac{5}{7}\right)^n+1}=\dfrac{4.0+7}{0+1}=7\)
\(\lim\dfrac{4^{n+1}+6^{n+2}}{5^n+8^n}=\lim\dfrac{8^n\left[4.\left(\dfrac{4}{8}\right)^n+36.\left(\dfrac{6}{8}\right)^n\right]}{8^n\left[\left(\dfrac{5}{8}\right)^n+1\right]}=\lim\dfrac{4.\left(\dfrac{4}{8}\right)^n+36\left(\dfrac{6}{8}\right)^n}{\left(\dfrac{5}{8}\right)^n+1}=\dfrac{0}{1}=0\)
7.
Do \(-1\le sin\left(n\pi\right)\le1\Rightarrow0\le1-sin\left(n\pi\right)\le2\)
\(\Rightarrow\dfrac{0}{n+1}\le\dfrac{1-sin\left(n\pi\right)}{n+1}\le\dfrac{2}{n+1}\)
Mà \(\lim\dfrac{0}{n+1}=\lim\dfrac{2}{n+1}=0\)
\(\Rightarrow\lim\dfrac{1-sin\left(n\pi\right)}{n+1}=0\) theo định lý kẹp
b.
\(sin10n+cos10n=\sqrt{2}sin\left(10n+\dfrac{\pi}{4}\right)\Rightarrow-\sqrt{2}\le sin10n+cos10n\le\sqrt{2}\)
\(\Rightarrow-\dfrac{\sqrt{2}}{n^2+2n}\le\dfrac{sin10n+cos10n}{n^2+2n}\le\dfrac{\sqrt{2}}{n^2+2n}\)
Mà \(\lim\left(-\dfrac{\sqrt{2}}{n^2+2n}\right)=\lim\left(\dfrac{\sqrt{2}}{n^2+2n}\right)=0\)
\(\Rightarrow\lim\dfrac{sin10n+cos10n}{n^2+2n}=0\)
8.
\(\lim\dfrac{n\sqrt{n}-1}{n+n\sqrt{n}}=\lim\dfrac{n\sqrt{n}\left(1-\dfrac{1}{n\sqrt{n}}\right)}{n\sqrt{n}\left(\dfrac{1}{\sqrt{n}}+1\right)}=\lim\dfrac{1-\dfrac{1}{n\sqrt{n}}}{\dfrac{1}{\sqrt{n}}+1}=\dfrac{1-0}{0+1}=1\)
\(\lim\dfrac{\sqrt[3]{n^2}+2}{n+\sqrt{n}}=\lim\dfrac{n\left(\dfrac{1}{\sqrt[3]{n}}+\dfrac{2}{n}\right)}{n\left(1+\dfrac{1}{\sqrt{n}}\right)}=\lim\dfrac{\dfrac{1}{\sqrt[3]{n}}+\dfrac{2}{n}}{1+\dfrac{1}{\sqrt{n}}}=\dfrac{0+0}{1+0}=0\)
\(\lim\dfrac{\sqrt[3]{n^3+3n^2+2}}{\sqrt{n^2-4n+5}}=\lim\dfrac{n\sqrt[3]{1+\dfrac{3}{n}+\dfrac{2}{n^3}}}{n\sqrt{1-\dfrac{4}{n}+\dfrac{5}{n^2}}}=\lim\dfrac{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{2}{n^3}}}{\sqrt{1-\dfrac{4}{n}+\dfrac{5}{n^2}}}=\dfrac{\sqrt[3]{1+0+0}}{\sqrt{1-0+0}}=1\)
9.
\(\lim\sqrt[3]{\dfrac{8n^2-3n}{n^2}}=\lim\sqrt[3]{\dfrac{n^2\left(8-\dfrac{3}{n}\right)}{n^2}}=\lim\sqrt[3]{8-\dfrac{3}{n}}=\sqrt[3]{8-0}=2\)
\(\lim\dfrac{2n^2-3n-1}{-n^2+2}=\lim\dfrac{n^2\left(2-\dfrac{3}{n}-\dfrac{1}{n^2}\right)}{n^2\left(-1+\dfrac{2}{n^2}\right)}=\lim\dfrac{2-\dfrac{3}{n}-\dfrac{1}{n^2}}{-1+\dfrac{2}{n^2}}=\dfrac{2}{-1}=-2\)
\(\lim\left(n-1-\sqrt{n^2+1}\right)=\lim\left(\dfrac{\left(n-\sqrt{n^2+1}\right)\left(n+\sqrt{n^2+1}\right)}{n+\sqrt{n^2+1}}-1\right)\)
\(=\lim\left(\dfrac{-1}{n+\sqrt{n^2+1}}-1\right)=0-1=-1\)
10.
\(\lim\dfrac{2n^3+2n-1}{3n^3-n+3}=\lim\dfrac{n^3\left(2+\dfrac{2}{n^2}-\dfrac{1}{n^3}\right)}{n^3\left(3-\dfrac{1}{n^2}+\dfrac{3}{n^3}\right)}=\lim\dfrac{2+\dfrac{2}{n^2}-\dfrac{1}{n^3}}{3-\dfrac{1}{n^2}+\dfrac{3}{n^3}}=\dfrac{2}{3}\)
\(\lim\dfrac{\sqrt{n^2+n}+n}{\sqrt{4n^2+1}+n-1}=\lim\dfrac{n\left(\sqrt{1+\dfrac{1}{n}}+1\right)}{n\left(\sqrt{4+\dfrac{1}{n^2}}+1-\dfrac{1}{n}\right)}=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}+1}{\sqrt{4+\dfrac{1}{n^2}}+1-\dfrac{1}{n}}=\dfrac{\sqrt{1+0}+1}{\sqrt{4+0}+1-1}=\dfrac{2}{3}\)
\(\lim\dfrac{4+2^n+3^{n+2}}{\left(-2\right)^{n+1}+5.3^n}=\lim\dfrac{3^n\left[4.\left(\dfrac{1}{3}\right)^n+\left(\dfrac{2}{3}\right)^n+3^2\right]}{3^n\left[-2.\left(-\dfrac{2}{3}\right)^n+5\right]}=\lim\dfrac{4\left(\dfrac{1}{3}\right)^n+\left(\dfrac{2}{3}\right)^n+9}{-2\left(-\dfrac{2}{3}\right)^n+5}=\dfrac{9}{5}\)
\(\lim\left(\sqrt{n^2+2n+3}-n\right)=\lim\dfrac{\left(\sqrt{n^2+2n+3}-n\right)\left(\sqrt{n^2+2n+3}+n\right)}{\sqrt{n^2+2n+3}+n}\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}==\dfrac{2}{2}=1\)
