Tìm các giới hạn sau: a) lim 2n-3/6n+1

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

Bài 2 trang 75 SBT Toán 11 Tập 1: Tìm các giới hạn sau:

a) $\text{lim}\frac{2n-3}{6n+1};$

b) $\text{lim}\frac{3n-1}{n^2+n};$

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

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

e) $\text{lim}\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right);$

g) $\text{lim}\frac{1}{\sqrt{n^2+n}-n}.$

Lời giải:

a) $\text{lim}\frac{2n-3}{6n+1}=\lim \frac{2-\frac{3}{n}}{6+\frac{1}{n}}=\frac{\lim 2-\lim \frac{3}{n}}{\lim 6+\lim \frac{1}{n}}$$=\frac{2-0}{6+0}=\frac{2}{6}=\frac{1}{3}.$

b) $\text{lim}\frac{3n-1}{n^2+n}=\lim \frac{\frac{3}{n}-\frac{1}{n^2}}{1+\frac{1}{n}}$$=\frac{\lim \frac{3}{n}-\lim \frac{1}{n^2}}{\lim 1+\lim \frac{1}{n}}=\frac{0-0}{1+0}=0.$

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

d) $\text{lim}\frac{4n+1}{\sqrt{n^2+3n}+n}=\text{lim}\frac{4+\frac{1}{n}}{\sqrt{1+\frac{3}{n}}+1}$$=\frac{4+\text{lim}\frac{1}{n}}{\sqrt{1+\text{lim}\frac{3}{n}}+1}=\frac{4}{1+1}=2$.

e) $\text{lim}\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)=\text{lim}\frac{\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}$

$=\text{lim}\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\text{lim}\frac{1}{\sqrt{1+\frac{1}{n}}+1}$

$=\frac{1}{\sqrt{1+\text{lim}\frac{1}{n}}+1}=\frac{1}{2}.$

g) $\text{lim}\frac{1}{\sqrt{n^2+n}-n}=\text{lim}\frac{\sqrt{n^2+n}+n}{\left(\sqrt{n^2+n}-n\right)\left(\sqrt{n^2+n}+n\right)}$

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