Bài 37. Tìm các đường tiệm cận của đồ thị mỗi hàm số sau:
a) \(y = x + \sqrt {{x^2} – 1} \) b) \(y = \sqrt {{x^2} – 4x + 3} \)
c) \(y = \sqrt {{x^2} + 4} \) d) \(y = {{{x^2} + x + 1} \over {{x^2} – 1}}\)
Gỉải
a) 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 \(x \to – \infty \))
b) 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 \(x \to + \infty \)).
* \(a = \mathop {\lim }\limits_{x \to – \infty } {y \over x} = \mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^2} – 4x + 3} } \over x} = \mathop {\lim }\limits_{x \to – \infty } {{ – x\sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} } \over x} = – \mathop {\lim }\limits_{x \to – \infty } \sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} = – 1\)
\(\eqalign{
& b = \mathop {\lim }\limits_{x \to – \infty } \left( {y + x} \right) = \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {{x^2} – 4x + 3} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } {{ – 4x + 3} \over {\sqrt {{x^2} – 4x + 3} – x}} = \mathop {\lim }\limits_{x \to – \infty } {{ – 4x + 3} \over { – x\sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} – x}} \cr
& \,\, = \,\,\,\mathop {\lim }\limits_{x \to – \infty } {{ – 4 + {3 \over x}} \over { – \sqrt {1 – {4 \over x} + {3 \over {{x^2}}}} – 1}} = {{ – 4} \over { – 2}} = 2 \cr} \)
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Tiệm cận xiên: \(y = -x + 2\) (khi \(x \to – \infty \)).
c) 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 \(x \to – \infty \))
d) 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.