Đáp án:
$\begin{array}{l}
1)a)\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{{x^2} + 2x}} - 1}}{{x + 2}}\\
= \dfrac{{\sqrt[3]{{{2^2} + 2.2}} - 1}}{{2 + 2}}\\
= \dfrac{{\sqrt[3]{8} - 1}}{4}\\
= \dfrac{1}{4}\\
b)\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 4x + 3}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {x - 3} \right)\left( {x - 1} \right)}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \left( {x - 1} \right)\\
= 3 - 1\\
= 2\\
c)\mathop {\lim }\limits_{x \to 6} \dfrac{{\sqrt {x + 3} - 3}}{{x - 6}}\\
= \mathop {\lim }\limits_{x \to 6} \dfrac{{x + 3 - 9}}{{\left( {x - 6} \right)\left( {\sqrt {x + 3} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 6} \dfrac{1}{{\sqrt {x + 3} + 3}}\\
= \dfrac{1}{{\sqrt {6 + 3} + 3}}\\
= \dfrac{1}{6}\\
2)a)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} - 2x + 5}}{{5 - 2{x^2}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 - \dfrac{2}{x} + \dfrac{5}{{{x^2}}}}}{{\dfrac{5}{{{x^2}}} - 2}}\\
= \dfrac{1}{{ - 2}}\\
= - \dfrac{1}{2}\\
b)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x + 1}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{2}{x} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{{{x^2}}}}} = 0\\
c)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 2{x^2} + x - 1}}{{3 - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 2x + 1 - \dfrac{1}{x}}}{{\dfrac{3}{x} - 1}}\\
= + \infty \\
d)\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x} - x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + 2x - {x^2}}}{{\sqrt {{x^2} + 2x} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x}}{{\sqrt {{x^2} + 2x} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{2}{{\sqrt {1 + \dfrac{2}{x}} + 1}}\\
= \dfrac{2}{{1 + 1}} = 1
\end{array}$
