Tính các giới hạn sau: a) lim (6n - 5)/3n; b) (-2n^2 - 6n + 2)/(8n^2 - 5n + 4)

Giải SBT Toán 11 Bài 1: Giới hạn của dãy số

Bài 9 trang 69 SBT Toán 11 Tập 1: Tính các giới hạn sau:

a) $\lim \frac{6n-5}{3n}$ ;

b) $\lim \frac{-2n^2-6n+2}{8n^2-5n+4}$ ;

c) $\lim \frac{n^3-5n+1}{3n^2-4n+2}$ ;

d) $\lim \frac{-4n+1}{9n^2-n+2}$ ;

e) $\lim \frac{\sqrt{4n^2+n+1}}{8n+3}$ ;

g) $\lim \frac{4^n+5^n}{{3.4}^n-{4.5}^n}$ .

Lời giải:

a) $\lim \frac{6n-5}{3n}=\lim \frac{n\left(6-\frac{5}{n}\right)}{3n}$$=\lim \frac{6-\frac{5}{n}}{3}=\frac{\lim \left(6-\frac{5}{n}\right)}{\lim 3}=\frac{6}{3}=2$ .

b) $\lim \frac{-2n^2-6n+2}{8n^2-5n+4}$ $=\lim \frac{n^2\left(-2-\frac{6}{n}+\frac{2}{n^2}\right)}{n^2\left(8-\frac{5}{n}+\frac{4}{n^2}\right)}$

$=\lim \frac{-2-\frac{6}{n}+\frac{2}{n^2}}{8-\frac{5}{n}+\frac{4}{n^2}}=\frac{\lim \left(-2-\frac{6}{n}+\frac{2}{n^2}\right)}{\lim \left(8-\frac{5}{n}+\frac{4}{n^2}\right)}=\frac{-2}{8}=-\frac{1}{4}$.

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

$=\frac{\lim \left(1-\frac{5}{n^2}+\frac{1}{n^3}\right)}{\lim \left(\frac{3}{n}-\frac{4}{n^2}+\frac{2}{n^3}\right)}=+\infty$ (do $\lim \left(1-\frac{5}{n^2}+\frac{1}{n^3}\right)=1$  và $\lim \left(\frac{3}{n}-\frac{4}{n^2}+\frac{2}{n^3}\right)=0$  ).

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

$=\frac{\lim \left(\frac{-4}{n}+\frac{1}{n^2}\right)}{\lim \left(9-\frac{1}{n}+\frac{2}{n^2}\right)}=\frac{0}{9}=0$ .

e) $\lim \frac{\sqrt{4n^2+n+1}}{8n+3}$ $=\lim \frac{\sqrt{n^2\left(4+\frac{1}{n}+\frac{1}{n^2}\right)}}{n\left(8+\frac{3}{n}\right)}$ $=\lim \frac{\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}}{8+\frac{3}{n}}$

$=\frac{\lim \left(\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}\right)}{\lim \left(8+\frac{3}{n}\right)}=\frac{\sqrt{\lim \left(4+\frac{1}{n}+\frac{1}{n^2}\right)}}{\lim \left(8+\frac{3}{n}\right)}=\frac{\sqrt{4}}{8}=\frac{2}{8}=\frac{1}{4}$.

g) $\lim \frac{4^n+5^n}{{3.4}^n-{4.5}^n}$$=\lim \frac{5^n\left(\frac{4^n}{5^n}+1\right)}{5^n\left(\frac{{3.4}^n}{5^n}-4\right)}$$=\lim \frac{{\left(\frac{4}{5}\right)}^n+1}{3.{\left(\frac{4}{5}\right)}^n-4}$