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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