Bài 2 trang 65 SGK Toán 11 Cánh diều Tập 1 - CD

a) \(lim\) $\frac{5n+1}{2n}$ = \(lim\) $\left(\frac{5}{2}+\frac{1}{2n}\right)=\mathrm{\backslash (lim\backslash ) }\frac{5}{2}+\mathrm{\backslash (lim\backslash ) }\frac{1}{2n}=\frac{5}{2}$.

b) \(\lim \frac{6{{n}^{2}}+8n+1}{5{{n}^{2}}+3}=\lim \frac{6+\frac{8}{n}+\frac{1}{{{n}^{2}}}}{5+\frac{3}{{{n}^{2}}}}=\frac{\lim \left( 6+\frac{8}{n}+\frac{1}{{{n}^{2}}} \right)}{\lim \left( 5+\frac{3}{{{n}^{2}}} \right)}=\frac{6}{5}\)

c) \(\lim \frac{\sqrt{{{n}^{2}}+5n+3}}{6n+2}=\lim \frac{\sqrt{1+\frac{5}{n}+\frac{3}{{{n}^{2}}}}}{6+\frac{2}{n}}=\frac{\lim \sqrt{1+\frac{5}{n}+\frac{3}{{{n}^{2}}}}}{\lim \left( 6+\frac{2}{n} \right)}=\frac{1}{6}\)

d) \(\lim \left( 2-\frac{1}{{{3}^{n}}} \right)=\lim 2-\lim {{\left( \frac{1}{3} \right)}^{n}}=2-0=2\)

e) \(\lim \frac{{{3}^{n}}+{{2}^{n}}}{{{4.3}^{n}}}=\lim \frac{1+{{\left( \frac{2}{3} \right)}^{n}}}{4}=\frac{\lim \left( 1+{{\left( \frac{2}{3} \right)}^{n}} \right)}{\lim 4}=\frac{1}{4}\)

g) \(\lim \frac{2+\frac{1}{n}}{{{3}^{n}}}=\frac{\lim \left( 2+\frac{1}{n} \right)}{\lim \left( {{3}^{n}} \right)}=0\)

-- Mod Toán 11 HỌC247