\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^9} + 6x + 8}}{{4\left| {{x^9}} \right| - 2x + 3}}\)
A.\(\dfrac{1}{4}\).
B.\(- \dfrac{1}{4}\).
C.\(0.\)
D.\(+ \infty\)

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }}\)
A.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = \dfrac{2}{3}\).
B.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = \pm\infty\).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = 3\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = - 3\).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = \dfrac{2}{3}\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = - \infty\).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \pm \infty \).
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \pm 1\) .
C.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = - \infty \).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} \) không tồn tại.
\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = +\infty \).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = \dfrac{1}{2}\)
B.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = \dfrac{1}{2}\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = - \infty\).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = \dfrac{1}{2}\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = - 1\).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = 1\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = - 1\).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = + \infty \).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\) không tồn tại. \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
D.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \mp \infty \).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = + \infty \).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\) không tồn tại. \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
D.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \mp \infty \).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = 1 \),
B.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = + \infty \), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = - \infty \)
C.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = - \infty \),
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = + 1 \), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = - 1 \)

\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 2x} - \sqrt {{x^2} + 1} }}{{4{x^2} - 1}}\)
A.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 2x} - \sqrt {{x^2} + 1} }}{{4{x^2} - 1}} = -\infty\).
B.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 2x} - \sqrt {{x^2} + 1} }}{{4{x^2} - 1}} = +\infty\).
C.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 2x} - \sqrt {{x^2} + 1} }}{{4{x^2} - 1}} = 0\).
D.\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} + 2x} - \sqrt {{x^2} + 1} }}{{4{x^2} - 1}} = \dfrac{1}{4}\).

Thằn lằn bóng đuôi dài thường sống ở
A.dưới nước.
B.những chỗ ẩm ướt.
C.những nơi khô ráo
D.trên không

Màng nhĩ của thằn lằn nằm trong một hốc nhỏ bên đầu có ý nghĩa
A.Nghe tốt hơn
B.Bảo vệ màng nhĩ và hướng các dao động âm thanh vào màng nhĩ.
C.giúp chúng giữ thăng bằng
D.Cả 3 phương án trên