Bài 37 trang 36 SGK giải tích 12 nâng cao

LG a

\(y = x + \sqrt {{x^2} – 1} \)

Lời giải chi tiết:

TXĐ: \(D = \left( { – \infty ; – 1} \right] \cup \left[ {1; + \infty } \right)\)
* \(a = \mathop {\lim }\limits_{x \to  + \infty } {y \over x} \) \(= \mathop {\lim }\limits_{x \to  + \infty } \left( {1 + {{\sqrt {{x^2} – 1} } \over x}} \right) \) \(= \mathop {\lim }\limits_{x \to  + \infty } \left( {1 + \sqrt {1 – {1 \over {{x^2}}}} } \right) = 2\)
\(b = \mathop {\lim }\limits_{x \to  + \infty } \left( {y – 2x} \right) \) \(= \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} – 1}  – x} \right) \) \(= \mathop {\lim }\limits_{x \to  + \infty } {{ – 1} \over {\sqrt {{x^2} – 1}  + x}} = 0\)
Ta có tiệm cận xiên \(y = 2x\) (khi \(x \to  + \infty \))
* \(\mathop {\lim }\limits_{x \to  – \infty } y = \mathop {\lim }\limits_{x \to  – \infty } \left( {x + \sqrt {{x^2} – 1} } \right) \) \(= \mathop {\lim }\limits_{x \to  – \infty } {{ – 1} \over {\sqrt {{x^2} – 1}  – x}} = 0\)

Ta có tiệm cận ngang \(y = 0\) (khi \(x \to  – \infty \))

LG b

\(y = \sqrt {{x^2} – 4x + 3} \)

Lời giải chi tiết:

TXĐ: \(D = \left( { – \infty ;1} \right] \cup \left[ {3; + \infty } \right)\)
* \(a = \mathop {\lim }\limits_{x \to  + \infty } {y \over x} \) \(= \mathop {\lim }\limits_{x \to  + \infty } {{\sqrt {{x^2} – 4x + 3} } \over x} \) \(= \mathop {\lim }\limits_{x \to  + \infty } \sqrt {1 – {4 \over x} + {3 \over {{x^2}}}}  = 1\)
\(b = \mathop {\lim }\limits_{x \to  + \infty } \left( {y – x} \right) \) \(= \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} – 4x + 3}  – x} \right)\) \( = \mathop {\lim }\limits_{x \to  + \infty } {{ – 4x + 3} \over {\sqrt {{x^2} – 4x + 3}  + x}} \) \(= \mathop {\lim }\limits_{x \to  + \infty } {{ – 4 + {3 \over x}} \over {\sqrt {1 – {4 \over x} + {3 \over {{x^2}}}}  + 1}} =  – 2\)
Ta có tiệm cận xiên \(y = x -2\) (khi \(x \to  + \infty \)).
* \(a = \mathop {\lim }\limits_{x \to  – \infty } {y \over x} \) \(= \mathop {\lim }\limits_{x \to  – \infty } {{\sqrt {{x^2} – 4x + 3} } \over x} \) \(= \mathop {\lim }\limits_{x \to  – \infty } {{ – x\sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} } \over x} \) \(=  – \mathop {\lim }\limits_{x \to  – \infty } \sqrt {1 – {4 \over x} + {3 \over {{x^2}}}}  =  – 1\)

\(\eqalign{
& b = \mathop {\lim }\limits_{x \to – \infty } \left( {y + x} \right) \cr&= \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} – 4x + 3} + x} \right) \cr&= \mathop {\lim }\limits_{x \to – \infty } {{ – 4x + 3} \over {\sqrt {{x^2} – 4x + 3} – x}} \cr&= \mathop {\lim }\limits_{x \to – \infty } {{ – 4x + 3} \over { – x\sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} – x}} \cr 
& = \mathop {\lim }\limits_{x \to – \infty } {{ – 4 + {3 \over x}} \over { – \sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} – 1}}\cr&  = {{ – 4} \over { – 2}} = 2 \cr} \)

Tiệm cận xiên: \(y = -x + 2\) (khi \(x \to  – \infty \)).

LG c

\(y = \sqrt {{x^2} + 4} \)

Lời giải chi tiết:

TXD: \(D =\mathbb R\)
* \(a = \mathop {\lim }\limits_{x \to  + \infty } {y \over x} \) \(= \mathop {\lim }\limits_{x \to  + \infty } \sqrt {1 + {4 \over {{x^2}}}}  = 1\)
\(b = \mathop {\lim }\limits_{x \to  + \infty } \left( {y – x} \right)  \) \(= \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + 4}  – x} \right)  \) \(= \mathop {\lim }\limits_{x \to  + \infty } {4 \over {\sqrt {{x^2} + 4}  + x}} = 0\)
Tiệm cận xiên \(y = x\) (khi \(x \to  + \infty \))
* \(a = \mathop {\lim }\limits_{x \to  – \infty } {y \over x}  \) \(= \mathop {\lim }\limits_{x \to  – \infty }- \sqrt {1 + {4 \over {{x^2}}}}  =  – 1\)
\(b = \mathop {\lim }\limits_{x \to  – \infty } \left( {y + x} \right)  \) \(= \mathop {\lim }\limits_{x \to  – \infty } \left( {\sqrt {{x^2} + 4}  + x} \right)  \) \( = \mathop {\lim }\limits_{x \to  – \infty } {4 \over {\sqrt {{x^2} + 4}  – x}} = 0\)

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

LG d

\(y = {{{x^2} + x + 1} \over {{x^2} – 1}}\)

Lời giải chi tiết:

TXĐ: \(D =\mathbb R\backslash \left\{ { – 1;1} \right\}\)
* Vì \(\mathop {\lim }\limits_{x \to  + \infty } y = \mathop {\lim }\limits_{x \to  + \infty } {{1 + {1 \over x} + {1 \over {{x^2}}}} \over {1 – {1 \over {{x^2}}}}} = 1\)
Tiệm cận ngang: \(y = 1\) (khi \(x \to  – \infty \) và \(x \to  + \infty \))
* \(\mathop {\lim }\limits_{x \to {1^ + }} y = \mathop {\lim }\limits_{x \to {1^ + }} {{{x^2} + x + 1} \over {\left( {x – 1} \right)\left( {x + 1} \right)}} =  + \infty \) và \(\mathop {\lim }\limits_{x \to {1^ – }} y = \mathop {\lim }\limits_{x \to {1^ – }} {{{x^2} + x + 1} \over {\left( {x – 1} \right)\left( {x + 1} \right)}} =  – \infty \) nên \(x = 1\) là tiệm cận đứng.
Tương tự: \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} y =  – \infty \) và \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ – }} y =  + \infty \) nên \(x = -1\) là tiệm cận đứng.

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