Bài 3.1 trang 64

Đề bài

Tìm các giới hạn:

a, $lim\frac{3n+2}{4-n}$

b, $lim\frac{5n^2+2n-1}{2n^2+n+1}$

c, $lim\frac{\sqrt{n^2+4n+2}}{3n-1}$

d, $lim\frac{n+7}{4+n^2}$

e, $lim\frac{2^n-1}{5^n+1}$

Lời giải chi tiết

a, Ta có: $\frac{3n+2}{4-n}=\frac{3+\frac{2}{n}}{\frac{4}{n}-1}$

Vì lim 3= 3, lim $\frac{2}{n}$=0, lim$\frac{4}{n}$=0, lim 1=1 nên $lim(3+\frac{2}{n})=3$ và $lim(\frac{4}{n}-1)$= -1

Vậy $lim\frac{3n+2}{4-n}=-3$.

b, Ta có: $\frac{5n^2+2n-1}{2n^2+n+1}=\frac{5+\frac{2}{n}-\frac{1}{n^2}}{2+\frac{1}{n}+\frac{1}{n^2}}$

Vì lim 5= 5, lim 2=2, $lim\frac{2}{n}=0$, $lim\frac{1}{n}=0$, $lim\frac{1}{n^2}=0$ nên $lim(5+\frac{2}{n}-\frac{1}{n^2})=5$ và $lim(2+\frac{1}{n}+\frac{1}{n^2})=2$.

Vậy $lim\frac{5n^2+2n-1}{2n^2+n+1}=\frac{5}{2}$.

c, Ta có: $\frac{\sqrt{n^2+4n+2}}{3n-1}=\frac{\frac{\sqrt{n^2+4n+2}}{n}}{\frac{3n-1}{n}}=\frac{\sqrt{\frac{n^2+4n+2}{n^2}}}{3-\frac{1}{n}}$=$\frac{\sqrt{1+\frac{4}{n}+\frac{2}{n^2}}}{3-\frac{1}{n}}$

Vì lim 1=1, lim 3=3, $lim\frac{4}{n}=0$, $lim\frac{2}{n^2}=0$, $lim\frac{1}{n}=0$ nên $lim\sqrt{1+\frac{4}{n}+\frac{2}{n^2}}=lim\sqrt{1}=1$ và $lim(3-\frac{1}{n})=3$

Vậy $lim\frac{\sqrt{n^2+4n+2}}{3n-1}=\frac{1}{3}$

d, Ta có: $\frac{n+7}{4+n^2}=\frac{\frac{1}{n}+\frac{7}{n^2}}{\frac{4}{n^2}+1}$

Vì lim 1=1, $lim\frac{1}{n}=0$; $lim\frac{7}{n^2}=0$; $lim\frac{4}{n^2}=0$ nên $lim(\frac{1}{n}+\frac{7}{n^2})=0$ và $lim(\frac{4}{n^2}+1)=1$

Vậy $lim\frac{n+7}{4+n^2}=0$.

e, Ta có: $\frac{2^n-1}{5^n+1}=\frac{{\left(\frac{2}{5}\right)}^n-\frac{1}{5^n}}{1+\frac{1}{5^n}}$

Vì lim 1=1 , $lim(\frac{2}{5}{)}^n=0$, $lim\frac{1}{5^n}=0$ nên $lim\left[{\left(\frac{2}{5}\right)}^n-\frac{1}{5^n}\right]=0$ và $lim\left(1+\frac{1}{5^n}\right)=1$

Vậy $lim\frac{2^n-1}{5^n+1}=0$.