Tính các giới hạn sau: a) lim (2n - 4)/5; b) lim (1+1/2n)/2n

Giải SBT Toán 11 Bài tập cuối chương 3

Bài 42 trang 83 SBT Toán 11 Tập 1: Tính các giới hạn sau:

a) $\lim \frac{2n-4}{5}$ ;                                   b) $\lim \frac{1+\frac{1}{2n}}{2n}$ ;

c) $\lim \left(2+\frac{7}{4^n}\right)$ ;                                 d) $\lim \frac{-4n^2-3}{2n^2-n+5}$ ;

e) $\lim \frac{\sqrt{9n^2+2n+1}}{n-5}$ ;                        g) $\lim \frac{3^n+{4.9}^n}{{3.4}^n+9^n}$ .

Lời giải:

a) Vì lim(2n – 4) = +∞ và lim5 = 5 > 0 nên $\lim \frac{2n-4}{5}=+\infty$ .                              

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

c)

$=\lim 2+\lim 7.\lim {\left(\frac{1}{4}\right)}^n=2+7.0=2$.          

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

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

e) $\lim \frac{\sqrt{9n^2+2n+1}}{n-5}$ $=\lim \frac{\sqrt{n^2\left(9+\frac{2}{n}+\frac{1}{n^2}\right)}}{n-5}$ $=\lim \frac{n\sqrt{9+\frac{2}{n}+\frac{1}{n^2}}}{n\left(1-\frac{5}{n}\right)}$

$=\lim \frac{\sqrt{9+\frac{2}{n}+\frac{1}{n^2}}}{1-\frac{5}{n}}=\frac{\lim \sqrt{9+\frac{2}{n}+\frac{1}{n^2}}}{\lim \left(1-\frac{5}{n}\right)}=\frac{\sqrt{9}}{1}=3$.