Bài 1.39 trang 18 SBT Giải tích 12 Nâng cao

LG a

$y = \sqrt {{x^2} - x + 1}$

Lời giải chi tiết:

Ta có :                

$a = \mathop {\lim }\limits_{x \to  + \infty } {y \over x} = \mathop {\lim }\limits_{x \to  + \infty } {{\sqrt {{x^2} - x + 1} } \over x}$

$= \mathop {\lim }\limits_{x \to  + \infty } {{x\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} } \over x}$

$= \mathop {\lim }\limits_{x \to  + \infty } \sqrt {1 - {1 \over x} + {1 \over {{x^2}}}}  = 1$

$\eqalign{& b = \mathop {\lim }\limits_{x \to  + \infty } (y - x) \cr&= \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} - x + 1}  - x} \right)  \cr &  = \mathop {\lim }\limits_{x \to  + \infty } {{ - x + 1} \over {\sqrt {{x^2} - x + 1}  + x}}  \cr &  = \mathop {\lim }\limits_{x \to  + \infty } {{ - 1 + {1 \over x}} \over {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}}  + 1}} =  - {1 \over 2} \cr}$

Đường thẳng $y = x - {1 \over 2}$ là tiệm cận xiên của đồ thị (khi $x \to  + \infty$)

$\eqalign{& a = \mathop {\lim }\limits_{x \to  - \infty } {y \over x} = \mathop {\lim }\limits_{x \to  - \infty } {{\sqrt {{x^2} - x + 1} } \over x} \cr&= \mathop {\lim }\limits_{x \to  - \infty } {{ - x\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} } \over x}  \cr &  = \mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} } \right) =  - 1 \cr}$

$\eqalign{& b = \mathop {\lim }\limits_{x \to  - \infty } (y + x)\cr& = \mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} - x + 1}  + x} \right) \cr&= \mathop {\lim }\limits_{x \to  - \infty } {{ - x + 1} \over {\sqrt {{x^2} - x + 1}  - x}}  \cr &  = \mathop {\lim }\limits_{x \to  - \infty } {{ - x + 1} \over { - x\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}}  - x}} \cr&= \mathop {\lim }\limits_{x \to  - \infty } {{ - 1 + {1 \over x}} \over { - \sqrt {1 - {1 \over x} + {1 \over {{x^2}}}}  - 1}} = {1 \over 2} \cr}$             

Đường thẳng $y =- x + {1 \over 2}$ là tiệm cận xiên của đồ thị (khi $x \to  - \infty$)

LG b

$y = x + \sqrt {{x^2} + 2x}$

Lời giải chi tiết:

$\begin{array}{l} a = \mathop {\lim }\limits_{x \to + \infty } \frac{y}{x} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + \sqrt {{x^2} + 2x} }}{x}\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {1 + \sqrt {1 + \frac{2}{x}} } \right) = 2\\ b = \mathop {\lim }\limits_{x \to + \infty } \left( {y - 2x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {x + \sqrt {{x^2} + 2x} - 2x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x} - x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{2x}}{{\sqrt {{x^2} + 2x} + x}}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{2}{{\sqrt {1 + \frac{2}{x}} + 1}} = 1\\ \Rightarrow a = 2,b = 1 \end{array}$

Tiệm cận xiên: y = 2x + 1 (khi $x \to  + \infty$)

$\begin{array}{l} \mathop {\lim }\limits_{x \to - \infty } \left( {x + \sqrt {{x^2} + 2x} } \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2x}}{{x - \sqrt {{x^2} + 2x} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2x}}{{x - \left| x \right|\sqrt {1 + \frac{2}{x}} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2x}}{{x + x\sqrt {1 + \frac{2}{x}} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2}}{{1 + \sqrt {1 + \frac{2}{x}} }} = - 1 \end{array}$

Tiệm cận ngang: y = -1 (khi $x \to  - \infty$)

LG c

$y = \sqrt {{x^2} + 3}$

Lời giải chi tiết:

$\begin{array}{l} a = \mathop {\lim }\limits_{x \to + \infty } \frac{y}{x} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 3} }}{x}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{3}{{{x^2}}}} }}{1} = 1\\ b = \mathop {\lim }\limits_{x \to + \infty } \left( {y - x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3} - x} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{\sqrt {{x^2} + 3} + x}} = 0\\ \Rightarrow a = 1,b = 0 \end{array}$

Tiệm cận xiên: y = x  (khi $x \to  + \infty$)

$\begin{array}{l} a = \mathop {\lim }\limits_{x \to - \infty } \frac{y}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 3} }}{x}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {1 + \frac{3}{{{x^2}}}} }}{x}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {1 + \frac{3}{{{x^2}}}} }}{x}\\ = \mathop {\lim }\limits_{x \to - \infty } \left( { - \sqrt {1 + \frac{3}{{{x^2}}}} } \right) = - 1\\ b = \mathop {\lim }\limits_{x \to - \infty } \left( {y + x} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3} + x} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{3}{{\sqrt {{x^2} + 3} - x}} = 0\\ \Rightarrow a = - 1,b = 0 \end{array}$

Tiệm cận xiên: y = -x (khi $x \to  - \infty$)

LG d

$y = x + {2 \over {\sqrt x }}$

Lời giải chi tiết:

$\mathop {\lim }\limits_{x \to {0^ + }} y = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x + \frac{2}{{\sqrt x }}} \right) =  + \infty$

Tiệm cận đứng: x = 0 (khi $x \to {0^ + }$)

$\mathop {\lim }\limits_{x \to  + \infty } \left( {y - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{2}{{\sqrt x }} = 0$

Tiệm cận xiên: y = x (khi $x \to  + \infty$)

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