Giải bài 10 Ôn tập cuối năm

Giải bài 10 Ôn tập cuối năm.

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

$= lim \frac{(1 + \frac{1}{n}) {(\frac{3}{n} - 2)}^{2}}{1 + \frac{1}{n^{3}}}$

$= \frac{(1 + 0) {(0 - 2)}^{2}}{1 + 0} = 4$

b. Ta có:

$\frac{1}{n^{2} + 1} + \frac{2}{n^{2} + 1} + \frac{3}{n^{2} + 1} + . . . + \frac{n - 1}{n^{2} + 1}$

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

$= \frac{\frac{n (n - 1)}{2}}{n^{2} + 1}$

$= \frac{n^{2} - n}{2 (n^{2} + 1)}$

$\Rightarrow lim (\frac{1}{n^{2} + 1} + \frac{2}{n^{2} + 1} + \frac{3}{n^{2} + 1} + . . . + \frac{n - 1}{n^{2} + 1})$

$= l i m \frac{n^{2} - n}{2 (n^{2} + 1)}$

$= lim \frac{n^{2} (1 - \frac{1}{n})}{2 n^{2} (1 + \frac{1}{n^{2}})}$

$= lim \frac{1 - \frac{1}{n}}{2 (1 + \frac{1}{n^{2}})} = \frac{1}{2}$

c. $lim \frac{\sqrt{4 n^{2} + 1} + n}{2 n + 1}$

$= lim \frac{n . \sqrt{4 + \frac{1}{n^{2}}} + n}{2 n + 1}$

$= lim \frac{n . (\sqrt{4 + \frac{1}{n^{2}}} + 1)}{n (2 + \frac{1}{n})}$

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

$= \frac{2 + 1}{2} = \frac{3}{2}$

d. $lim \sqrt{n} (\sqrt{n - 1} - \sqrt{n})$

$= lim \frac{\sqrt{n} (\sqrt{n - 1} - \sqrt{n}) (\sqrt{n - 1} + \sqrt{n})}{\sqrt{n - 1} + \sqrt{n}}$

$= lim \frac{\sqrt{n} [(n - 1) - n]}{\sqrt{n - 1} + \sqrt{n}}$

$= lim \frac{- \sqrt{n}}{\sqrt{n} (\sqrt{1 - \frac{1}{n}} + 1)}$

$= lim \frac{- 1}{\sqrt{1 - \frac{1}{n}} + 1} = - \frac{1}{2}$