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

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.arrived
B.were singing
C.on the stage
D.when

A.as well as
B.friends
C.go
D.at

A.interested
B.which
C.at
D.last Sunday

A.that
B.that
C.contributing
D.services

A.importance
B.economy
C.educated
D.the whole country

A.What I told her
B.were
C.to
D.most of