Đáp án:
$\begin{array}{l}
4)\mathop {\lim }\limits_{ \to - \infty } \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{ \to - \infty } \frac{{\frac{1}{x} + 3}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} = - \frac{3}{{\sqrt 2 }} = \frac{{ - 3\sqrt 2 }}{2}\\
5)I = \mathop {\lim }\limits_{x \to 3} \frac{{\left| {x - 3} \right|}}{{x - 3}}\\
+ khi:x \to {3^ + } \Rightarrow I = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{x - 3}} = 1\\
+ khi:x \to {3^ - } \Rightarrow I = \mathop {\lim }\limits_{x \to 3} \frac{{3 - x}}{{x - 3}} = - 1\\
\Rightarrow kxd\\
6)\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}} = \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {{x^2} - 2x + 4} \right)}}{{{x^2} + 1}}\\
= \frac{{ - 2.12}}{{ - 4 + 1}} = 8\\
7)\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^3} - {x^2}} }}{{\sqrt {x - 1} + 1 - x}}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {x - 1} }}{{\left( {1 - \sqrt {x - 1} } \right).\sqrt {x - 1} }}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \frac{x}{{1 - \sqrt {x - 1} }}\\
= 1
\end{array}$
