A.The more
B.we use
C.the more better
D.will become

A.do housewives have to
B.get it done
C.for
D.cleaners

A.instead of
B.electronics
C.tend to toss
D.In

A.protect
B.forbidding
C.to restrict
D.and creating

A.with
B.were
C.when
D.started

A.was
B.play
C.making
D.pictures

A.\(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } }  - \sqrt {3x} } \right) = \dfrac{1}{2}\) \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } }  - \sqrt {3x} } \right)\) không tồn tại.
B.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } }  - \sqrt {3x} } \right) = \dfrac{1}{2}\)
C.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } }  - \sqrt {3x} } \right) = -\dfrac{1}{2}\)
D.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {3x + \sqrt {3x + \sqrt {3x} } }  - \sqrt {3x} } \right)\) không tồn tại

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

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.\(\mathop {\lim }\limits_{x \to  \pm \infty } \left( {\sqrt {4{x^2} + 2x + x}  - 1} \right) =  \pm \infty \)

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