Bài 1.38 trang 18 SBT Giải tích 12 Nâng cao

LG a

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

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {0^ + }} y = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{2{x^2} + 1}}{{{x^2} – 2x}}\\
\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{2{x^2} + 1}}{{x – 2}}.\frac{1}{x}} \right) = – \infty \\
\mathop {\lim }\limits_{x \to {0^ – }} y = + \infty \\
\mathop {\lim }\limits_{x \to {2^ + }} y = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{2{x^2} + 1}}{{{x^2} – 2x}}\\
\mathop {\lim }\limits_{x \to {2^ + }} \left( {\frac{{2{x^2} + 1}}{x}.\frac{1}{{x – 2}}} \right) = + \infty \\
\mathop {\lim }\limits_{x \to {2^ – }} y = – \infty \\
\mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } \frac{{2{x^2} + 1}}{{{x^2} – 2x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{2 + \frac{1}{{{x^2}}}}}{{1 – \frac{2}{x}}} = 2\\
\mathop {\lim }\limits_{x \to – \infty } y = 2
\end{array}\)

Vậy:

Đường thẳng x = 0 là tiệm cận đứng của đồ thị.

Đường thẳng x = 2  là tiệm cận đứng của đồ thị.

Đường thẳng y = 2 là tiệm cận ngang của đồ thị.

LG b

\(y = {x \over {1 – {x^2}}}\)

Lời giải chi tiết:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} y = \mathop {\lim }\limits_{x \to {1^ + }} \frac{x}{{1 – {x^2}}}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{x}{{1 + x}}.\frac{1}{{1 – x}}} \right) = – \infty \\
\mathop {\lim }\limits_{x \to {1^ – }} y = + \infty
\end{array}\)

Tiệm cận đứng: x = 1.

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to – {1^ + }} y = \mathop {\lim }\limits_{x \to – {1^ + }} \frac{x}{{1 – {x^2}}}\\
= \mathop {\lim }\limits_{x \to – {1^ + }} \left( {\frac{x}{{1 – x}}.\frac{1}{{1 + x}}} \right) = – \infty \\
\mathop {\lim }\limits_{x \to – {1^ – }} y = + \infty
\end{array}\)

Tiệm cận đứng: x = -1.

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } \frac{x}{{1 – {x^2}}} = 0\\
\mathop {\lim }\limits_{x \to – \infty } y = 0
\end{array}\)

Tiệm cận ngang: y = 0.

LG c

\(y = {{{x^3}} \over {{x^2} – 1}}\)

Lời giải chi tiết:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} y = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^3}}}{{{x^2} – 1}}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{{x^3}}}{{x + 1}}.\frac{1}{{x – 1}}} \right) = + \infty \\
\mathop {\lim }\limits_{x \to {1^ – }} y = – \infty
\end{array}\)

Tiệm cận đứng: x = 1 (khi \(x \to {1^ + }\) và \(x \to {1^ – }\))

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to – {1^ + }} y = \mathop {\lim }\limits_{x \to – {1^ + }} \frac{{{x^3}}}{{{x^2} – 1}}\\
= \mathop {\lim }\limits_{x \to – {1^ + }} \left( {\frac{{{x^3}}}{{x – 1}}.\frac{1}{{x + 1}}} \right) = + \infty \\
\mathop {\lim }\limits_{x \to – {1^ – }} y = – \infty
\end{array}\)

Tiệm cận đứng: x = -1.

\(\begin{array}{l}
y = \frac{{{x^3}}}{{{x^2} – 1}} = x + \frac{x}{{{x^2} – 1}}\\
\mathop {\lim }\limits_{x \to + \infty } \left( {y – x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{x}{{{x^2} – 1}} = 0\\
\mathop {\lim }\limits_{x \to – \infty } \left( {y – x} \right) = 0
\end{array}\)

Tiệm cận xiên: y = x.

LG d

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

Lời giải chi tiết:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ + }} y = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt x }}{{4 – {x^2}}}\\
= \mathop {\lim }\limits_{x \to {2^ + }} \left( {\frac{{\sqrt x }}{{2 + x}}.\frac{1}{{2 – x}}} \right) = – \infty \\
\mathop {\lim }\limits_{x \to {2^ – }} y = + \infty
\end{array}\)

Tiệm cận đứng: x = 2.

\(\mathop {\lim }\limits_{x \to  + \infty } y = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt x }}{{4 – {x^2}}} = 0\)

Tiệm cận ngang: y = 0.

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