Bài tập 6 trang 76 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) $1+3n-n^2=n^2\left(\frac{1}{n^2}+\frac{3}{n}-1.\right)$

Ta có: limn² = +∞ và $\lim \left(\frac{1}{n^2}+\frac{3}{n}-1\right)=0+0-1=-1.$

Suy ra $lim\left(1+3n–n^2\right)=\lim \left[n^2\left(\frac{1}{n^2}+\frac{3}{n}-1\right)\right]=-\infty .$

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

Ta có: limn² = +∞ và $\lim \frac{1+\frac{3}{n^2}}{2-\frac{1}{n}}=\frac{1}{2}.$

Suy ra $\text{lim}\frac{n^3+3n}{2n-1}=\text{lim}\left[n^2\cdot \frac{1+\frac{3}{n^2}}{2-\frac{1}{n}}\right]=+\infty .$

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

Ta có: limn = +∞ và $\lim \left(\sqrt{1-\frac{1}{n}}+1\right)=2.$

Suy ra $\text{lim}\left(\sqrt{n^2-n}+n\right)=\text{lim}\left[n\left(\sqrt{1-\frac{1}{n}}+1\right)\right]=+\infty .$

d) $3^{n+1}-5^n=5^n\left(\frac{3^{n+1}}{5^n}-1\right)=5^n\left[3\cdot {\left(\frac{3}{5}\right)}^n-1.\right]$

Ta có: lim5ⁿ = +∞ và $\lim \left[3\cdot {\left(\frac{3}{5}\right)}^n-1\right]=-1.$

Suy ra $\text{lim}\left(3^{n+1}-5^n\right)=\text{lim}\left\{5^n\cdot \left[3\cdot {\left(\frac{3}{5}\right)}^n-1\right]\right\}=-\infty .$

-- Mod Toán 11 HỌC247