Giải bài 13 Ôn tập cuối năm

Giải bài 13 Ôn tập cuối năm.

a) $\underset{x\to -2}{lim}\frac{6-3x}{\sqrt{2x^2+1}}=\frac{6-3(-2)}{\sqrt{2{(-2)}^2+1}}=\frac{12}{3}=4$

b)$\underset{x\to 2}{lim}\frac{x-\sqrt{3x-2}}{x^2-4}$

$=\underset{x\to 2}{lim}\frac{(x-\sqrt{3x-2})(x+\sqrt{3x-2})}{(x^2-4)(x+\sqrt{3x-2})}$

$=\underset{x\to 2}{lim}\frac{x^2-3x+2}{(x^2-4)(x+\sqrt{3x-2})}$

$=\underset{x\to 2}{lim}\frac{(x-2)(x-1)}{(x-2)(x+2)(x+\sqrt{3x-2)}}$

$=\underset{x\to 2}{lim}\frac{x-1}{(x+2)(x+\sqrt{3x-2})}$

$=\frac{2-1}{(2+2)(2+\sqrt{3.2-2})}=\frac{1}{16}$

c) Ta có:

• $\underset{x\to 2^{+}}{lim}(x^2-3x+1)=4-6+1=-1$ 
• $\{\begin{array}{c}x-2>0\hfill \\ \underset{x\to 2^{+}}{lim}(x-2)=0\hfill \end{array}$

Vậy $\underset{x\to 2^{+}}{lim}\frac{x^2-3x+1}{x-2}=-\mathrm{\infty }$

d) Ta có:

$\underset{x\to 1^{-}}{lim}(x+x^2+...+x^n-\frac{n}{1-x})=-\mathrm{\infty }$

$\{\begin{array}{c}1-x>0,\mathrm{\forall }x<1\hfill \\ \underset{x\to 1^{-}}{lim}(1-x)=0\hfill \end{array}$

$\Rightarrow \underset{x\to 1^{-}}{lim}\frac{n}{1-x}=+\mathrm{\infty }$

Vậy $\underset{x\to 1^{-}}{lim}(x+x^2+...+x^n-\frac{n}{1-x})=-\mathrm{\infty }$

e)$\underset{x\to +\mathrm{\infty }}{lim}\frac{2x-1}{x+3}=\underset{x\to +\mathrm{\infty }}{lim}\frac{x(2-\frac{1}{x})}{x(1+\frac{3}{x})}$

$=\underset{x\to +\mathrm{\infty }}{lim}\frac{2-\frac{1}{x}}{1+\frac{3}{x}}=2$

f) $\underset{x\to -\mathrm{\infty }}{lim}\frac{x+\sqrt{4x^2-1}}{2-3x}$

$=\underset{x\to -\mathrm{\infty }}{lim}\frac{x+|x|\sqrt{4-\frac{1}{x^2}}}{2-3x}$

$=\underset{x\to -\mathrm{\infty }}{lim}\frac{x-x\sqrt{4-\frac{1}{x^2}}}{2-3x}$

$=\underset{x\to -\mathrm{\infty }}{lim}\frac{x(1-\sqrt{4-\frac{1}{x^2}})}{x(\frac{2}{x}-3)}$

$=\underset{x\to -\mathrm{\infty }}{lim}\frac{1-\sqrt{4-\frac{1}{x^2}}}{\frac{2}{x}-3}$

$=\frac{1-\sqrt{4}}{-3}=\frac{1}{3}$

g) $\underset{x\to -\mathrm{\infty }}{lim}(-2x^3+x^2-3x+1)$

$=\underset{x\to -\mathrm{\infty }}{lim}{x}^3(-2+\frac{1}{x}-\frac{3}{x^2}+\frac{1}{x^3})$

$=+\mathrm{\infty }$