Tìm các giới hạn sau:

Giải

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

Vì lim 6-$\frac{1}{n}$ = lim6 -lim$\frac{1}{n}$= 6 

Và lim 3+$\frac{2}{n}$= lim 3 - lim$\frac{2}{n }$= 3 nên $\frac{6n-1}{3n+2}= lim\frac{6-{\displaystyle \frac{1}{n}}}{3+{\displaystyle \frac{2}{n}}} =\frac{6}{3}=2$

b)lim$\frac{3n^2+ n -5}{2n^2+1}$= lim$\frac{n^2\left(3+{\displaystyle \frac{1}{n}}-{\displaystyle \frac{5}{n^2}}\right)}{n^2\left(2+{\displaystyle \frac{1}{n^2}}\right)}= lim$ $\frac{3+{\displaystyle \frac{1}{n}}-{\displaystyle \frac{5}{n^2}}}{2+{\displaystyle \frac{1}{n^2}}}$= $\frac{lim 3 +lim{\displaystyle \frac{1}{n}}-lim{\displaystyle \frac{5}{n^2}}}{lim2 - lim{\displaystyle \frac{1}{n^2}}}$=$\frac{3}{2}$

c)$lim\frac{3^n+5.4^n}{4^n+2^n}$= lim$\frac{4^n\left[{\left({\displaystyle \frac{3}{4}}\right)}^n+5\right]}{4^n\left[1 - {\left({\displaystyle \frac{1}{2}}\right)}^n\right]}= lim\frac{{\left({\displaystyle \frac{3}{4}}\right)}^n+5}{1-{\left({\displaystyle \frac{1}{2}}\right)}^n} =5$

Vì lim ${\left(\frac{3}{4}\right)}^n=lim{\left(\frac{1}{2}\right)}^n$=0

d)lim$\frac{\sqrt{9n^2-n+1}}{4n-2}=lim\frac{n\sqrt{9-{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^2}}}}{n\left(4-{\displaystyle \frac{2}{n}}\right)}= lim\frac{\sqrt{9-{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^2}}}}{4-{\displaystyle \frac{2}{n}}}=\frac{3}{4}$