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khangle987
2 năm trước

A.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {4{x^2} + 2x + x}  - 1} \right) =  0\)
B.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {4{x^2} + 2x + x}  - 1} \right) =  - \infty \)
C.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {4{x^2} + 2x + x}  - 1} \right) =  + \infty \)
D.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {4{x^2} + 2x + x}  - 1} \right) =  \pm \infty \)
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A.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = + \infty \end{array}\)
B.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) =- \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = + \infty\end{array}\)
C.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = - \infty\end{array}\)
D.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) =- \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = - \infty\end{array}\)


A.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = + \infty \end{array}\)
B.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) =- \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = + \infty\end{array}\)
C.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = - \infty\end{array}\)
D.\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) =- \dfrac{3}{2}\\\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 3x + 1} - x} \right) = - \infty\end{array}\)

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