Bài 42 trang 83 SBT Toán 11 - Cánh diều: Tính các giới hạn sau: $\lim \frac{{2n - 4}}{5}$ $\lim \frac{{1 + \frac{1}{{2n}}}}{{2n}}$ c) $\lim \left( {2 + \frac{7}{{{4^n}}}} \right)$...

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) $\lim \frac{{ - 4{n^2} - 3}}{{2{n^2} - n + 5}}$

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

g) $\lim \frac{{{3^n} + {{4.9}^n}}}{{{{3.4}^n} + {9^n}}}$

Sử dụng các tính chất về giới hạn hàm số.

a) Ta có $\lim \left( {2n - 4} \right) = \lim \left[ {n\left( {2 - \frac{4}{n}} \right)} \right] = \lim n.\lim \left( {2 - \frac{4}{n}} \right) = 2\lim n = + \infty$

Suy ra $\lim \frac{{2n - 4}}{5} = + \infty$.

b) Ta có: $\lim \left( {1 + \frac{1}{{2n}}} \right) = \lim 1 + \lim \frac{1}{{2n}} = 1 + 0 = 1$ và $\lim 2n = + \infty$.

Suy ra $\lim \frac{{1 + \frac{1}{{2n}}}}{{2n}} = 0$.

c) Ta có $\lim \left( {2 + \frac{7}{{{4^n}}}} \right) = \lim 2 + \lim \frac{7}{{{4^n}}} = 2 + 0 = 2$.

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

$= \lim \frac{{ - 4 - \frac{3}{{{n^2}}}}}{{2 - \frac{1}{n} + \frac{5}{{{n^2}}}}} = \frac{{\lim \left( { - 4} \right) - \lim \frac{3}{{{n^2}}}}}{{\lim 2 - \lim \frac{1}{n} + \lim \frac{5}{{{n^2}}}}} = \frac{{ - 4 - 0}}{{2 - 0 + 0}} = - 2$

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

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

Do $\lim \left( {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} \right) = \lim 9 + \lim \frac{2}{n} + \lim \frac{1}{{{n^2}}} = 9 + 0 + 0 = 9$, ta suy ra:

$\lim \sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} = \sqrt 9 = 3$.

Mặt khác, $\lim \left( {1 - \frac{5}{n}} \right) = \lim 1 - \lim \frac{5}{n} = 1 - 0 = 1$

Suy ra $\lim \frac{{\sqrt {9{n^2} + 2n + 1} }}{{n - 5}} = \lim \frac{{\sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{1 - \frac{5}{n}}} = \frac{3}{1} = 3$.

f) Ta có: $\lim \frac{{{3^n} + {{4.9}^n}}}{{{{3.4}^n} + {9^n}}} = \lim \frac{{\frac{{{3^n}}}{{{9^n}}} + 4}}{{3.\frac{{{4^n}}}{{{9^n}}} + 1}} = \frac{{\lim {{\left( {\frac{3}{9}} \right)}^n} + \lim 4}}{{\lim 3.\lim {{\left( {\frac{4}{9}} \right)}^n} + \lim 1}} = \frac{{0 + 4}}{{3.0 + 1}} = 4$