Tìm các đường tiệm cận của đồ thị các hàm số sau
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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