Cho hàm số \(y = f\left( x \right) = \dfrac{{\sqrt {{x^2} - 9} - x}}{{{{(x - 4)}^2}}}\).
Cho hàm số \(y = f\left( x \right) = \dfrac{{\sqrt {{x^2} - 9} - x}}{{{{(x - 4)}^2}}}\).
Đúng | Sai | |
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1) Tập xác định của hàm số \(y = f\left( x \right)\) là \(D = \left( { - \infty ; - 3\left] \cup \right[3;4} \right) \cup \left( {4; + \infty } \right)\). |
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2) \(\mathop {\lim }\limits_{x \to 5} f\left( x \right) = - 1\). | ||
3) \(\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = 0\). | ||
4) \(\mathop {\lim }\limits_{x \to - \infty } \left[ {\left( {2x - 1} \right)f\left( x \right)} \right] = 0\). |
Đáp án đúng là: 1Đ, 2Đ, 3Đ, 4S
\(y = f\left( x \right) = \dfrac{{\sqrt {{x^2} - 9} - x}}{{{{(x - 4)}^2}}}\)
a) Tập xác định là \(\left\{ \begin{array}{l}{x^2} - 9 \ge 0\\{\left( {x - 4} \right)^2} \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \le - 3\\x \ge 3\end{array} \right.\\x \ne 4\end{array} \right. \Leftrightarrow x \in \left( { - \infty ; - 3\left] \cup \right[3;4} \right) \cup \left( {4; + \infty } \right)\)
vậy \(D = \left( { - \infty ; - 3\left] \cup \right[3;4} \right) \cup \left( {4; + \infty } \right)\)
b) \(\mathop {\lim }\limits_{x \to 5} f\left( x \right) = \mathop {\lim }\limits_{x \to 5} \dfrac{{\sqrt {{x^2} - 9} - x}}{{{{(x - 4)}^2}}} = \dfrac{{\sqrt {{5^2} - 9} - 5}}{{{{(5 - 4)}^2}}} = - 1\)
c) \(\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} - 9} - x}}{{{{(x - 4)}^2}}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 9}}{{{{(x - 4)}^2}\left( {\sqrt {{x^2} - 9} + x} \right)}} = 0\)
d) \(\mathop {\lim }\limits_{x \to - \infty } \left[ {\left( {2x - 1} \right)f\left( x \right)} \right] = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {2x - 1} \right)\left( {\sqrt {{x^2} - 9} - x} \right)}}{{{{(x - 4)}^2}}}\) \( = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {2 - \dfrac{1}{x}} \right)\left( { - \sqrt {1 - \dfrac{9}{x}} - 1} \right)}}{{{{\left( {1 - \dfrac{4}{x}} \right)}^2}}} = - 4\)
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