Giải câu 3 bài 1: Giới hạn của dãy số

a) \(\lim \frac{6n - 1}{3n +2} = \lim\frac{n\left ( 6 - \frac{1}{n} \right )}{n\left ( 3 +\frac{2}{n} \right )} =\lim\frac{6 - \frac{1}{n}}{3 +\frac{2}{n}} = \frac{6-0}{3-0} = 2\).

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

\(=\lim \frac{3 +\frac{1}{n}-\frac{5}{n^{2}}}{2+\frac{1}{n^{2}}}= \frac{3}{2}\).

c) \(\lim \frac{3^{n}+5.4^{n}}{4^{n}+2^{n}}\)

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

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

\(=\frac{5}{1}= 5\)

d) \(\lim \frac{\sqrt{9n^{2}-n+1}}{4n -2}\)

\(=\lim \frac{\sqrt{{n^2}\left( {9 - {1 \over n} + {1 \over {{n^2}}}} \right)}}{n(4-\frac{2}{n})}\)

\(=\lim \frac{n\sqrt{\left( {9 - {1 \over n} + {1 \over {{n^2}}}} \right)}}{n(4-\frac{2}{n})}\)

\(=\lim \frac{\sqrt{9-\frac{1}{n}+\frac{1}{n^{2}}}}{4-\frac{2}{n}}\)

\(=\frac{\sqrt{9}}{4}=\frac{3}{4}\).