Câu 5 trang 142 Đại số và giải tích 11: Ôn tập chương IV - Giới hạn...

Bài 5. Tính các giới hạn sau

a) $\mathop {\lim }\limits_{x \to 2} {{x + 3} \over {{x^2} + x + 4}}$                                                      

b) $\mathop {\lim }\limits_{x \to  - 3} {{{x^2} + 5x + 6} \over {{x^2} + 3x}}$

c) $\mathop {\lim }\limits_{x \to {4^ - }} {{2x - 5} \over {x - 4}}$                                                              

d) $\mathop {\lim }\limits_{x \to  + \infty } ( - {x^3} + {x^2} - 2x + 1)$

e) $\mathop {\lim }\limits_{x \to  - \infty } {{x + 3} \over {3x - 1}}$                                                              

f) $\mathop {\lim }\limits_{x \to  - \infty } {{\sqrt {{x^2} - 2x + 4}  - x} \over {3x - 1}}$

a) $\mathop {\lim }\limits_{x \to 2} {{x + 3} \over {{x^2} + x + 4}} = {{2 + 3} \over {{2^2} + 2 + 4}} = {1 \over 2}$

b)

$\eqalign{ & \mathop {\lim }\limits_{x \to - 3} {{{x^2} + 5x + 6} \over {{x^2} + 3x}} = \mathop {\lim }\limits_{x \to - 3} {{(x + 2)(x + 3)} \over {x(x + 3)}} = \mathop {\lim }\limits_{x \to - 3} {{x + 2} \over x} \cr & = {{ - 3 + 2} \over { - 3}} = {1 \over 3} \cr}$

c) $\mathop {\lim }\limits_{x \to {4^ - }} {{2x - 5} \over {x - 4}}$

Ta có: 

$\mathop {\lim }\limits_{x \to {4^ - }} (2x - 5) = 3 > 0$(1)

$\left\{ \matrix{ x - 4 < 0,\forall x < 4 \hfill \cr \mathop {\lim }\limits_{x \to - 4} (x - 4) = 0 \hfill \cr} \right.$

(2)

Từ (1) và (2) suy ra: $\mathop {\lim }\limits_{x \to {4^ - }} {{2x - 5} \over {x - 4}} =  - \infty$

d) $\mathop {\lim }\limits_{x \to  + \infty } ( - {x^3} + {x^2} - 2x + 1) = \mathop {\lim }\limits_{x \to  + \infty } {x^3}( - 1 + {1 \over x} - {2 \over {{x^2}}} + {1 \over {{x^3}}}) =  - \infty$

e) 

$\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{x + 3} \over {3x - 1}} = \mathop {\lim }\limits_{x \to - \infty } {{x(1 + {3 \over x})} \over {x(3 - {1 \over x})}} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{1 + {3 \over x}} \over {3 - {1 \over x}}} = {1 \over 3} \cr}$

 f) 

$\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^2} - 2x + 4} - x} \over {3x - 1}} = \mathop {\lim }\limits_{x \to - \infty } {{|x|\sqrt {1 - {2 \over x} + {4 \over {{x^2}}}} - x} \over {3x - 1}} \cr & \mathop {\lim }\limits_{x \to - \infty } {{ - x\sqrt {1 - {2 \over x} + {4 \over {{x^2}}}} - x} \over {x(3 - {1 \over x})}} = \mathop {\lim }\limits_{x \to - \infty } {{ - \sqrt {1 - {2 \over x} + {4 \over {{x^2}}}} - 1} \over {3 - {1 \over x}}} = {{ - 2} \over 3} \cr}$.