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 \(x \to + \infty \)).* \(a = \mathop {\lim }\limits_{x \to - \infty } {y \over x} \) \(= \mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^2} - 4x + 3} } \over x} \) \(= \mathop {\lim }\limits_{x \to - \infty } {{ - x\sqrt {1 - {4 \over x} + {3 \over {{x^2}}}} } \over x} \) \(= - \mathop {\lim }\limits_{x \to - \infty } \sqrt {1 - {4 \over x} + {3 \over {{x^2}}}} = - 1\)
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Tìm các đường tiệm cận của đồ thị mỗi hàm số sau:
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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