Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - x + 3}}{{2x - 1}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 1 + \dfrac{3}{x}}}{{2 - \dfrac{1}{x}}} = \dfrac{{ - 1 + 0}}{{2 - 0}} = - \dfrac{1}{2}\\
2,\\
\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^3} + 3x - 4}}{{ - {x^3} - {x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 + \dfrac{3}{{{x^2}}} - \dfrac{4}{{{x^3}}}}}{{ - 1 - \dfrac{1}{x} + \dfrac{1}{{{x^3}}}}} = \dfrac{{2 + 0 - 0}}{{ - 1 - 0 + 0}} = - 2\\
3,\\
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} - x + 5} }}{{2x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2}\left( {1 - \dfrac{1}{x} + \dfrac{5}{{{x^2}}}} \right)} }}{{2x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left| x \right|.\sqrt {1 - \dfrac{1}{x} + \dfrac{5}{{{x^2}}}} }}{{2x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - x.\sqrt {1 - \dfrac{1}{x} + \dfrac{5}{{{x^2}}}} }}{{2x - 1}}\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \sqrt {1 - \dfrac{1}{x} + \dfrac{5}{{{x^2}}}} }}{{2 - \dfrac{1}{x}}} = \dfrac{{ - \sqrt {1 - 0 + 0} }}{{2 - 0}} = - \dfrac{1}{2}\\
4,\\
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2} - 3x} + 2x}}{{3x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^2}\left( {1 - \dfrac{3}{x}} \right)} + 2x}}{{3x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left| x \right|\sqrt {1 - \dfrac{3}{x}} + 2x}}{{3x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - x\sqrt {1 - \dfrac{3}{x}} + 2x}}{{3x - 1}}\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \sqrt {1 - \dfrac{3}{x}} + 2}}{{3 - \dfrac{1}{x}}} = \dfrac{{ - \sqrt 1 + 2}}{3} = \dfrac{1}{3}\\
5,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x + 3} - x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {\sqrt {{x^2} + 2x + 3} - x} \right)\left( {\sqrt {{x^2} + 2x + 3} + x} \right)}}{{\sqrt {{x^2} + 2x + 3} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {{x^2} + 2x + 3} \right) - {x^2}}}{{\sqrt {{x^2} + 2x + 3} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x + 3}}{{\sqrt {{x^2} + 2x + 3} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 + \dfrac{3}{x}}}{{\sqrt {1 + \dfrac{2}{x} + \dfrac{3}{{{x^2}}}} + 1}} = \dfrac{{2 + 0}}{{\sqrt {1 + 0 + 0} + 1}} = 1\\
6,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {2x - \sqrt {4{x^2} - x + 3} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {2x - \sqrt {4{x^2} - x + 3} } \right)\left( {2x + \sqrt {4{x^2} - x + 3} } \right)}}{{2x + \sqrt {4{x^2} - x + 3} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {2x} \right)}^2} - \left( {4{x^2} - x + 3} \right)}}{{2x + \sqrt {4{x^2} - x + 3} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x - 3}}{{2x + \sqrt {4{x^2} - x + 3} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 - \dfrac{3}{x}}}{{2 + \sqrt {4 - \dfrac{1}{x} + \dfrac{3}{{{x^2}}}} }} = \dfrac{{1 - 0}}{{2 + \sqrt {4 - 0 + 0} }} = \dfrac{1}{4}
\end{array}\)
\(\begin{array}{l}
7,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + x - 1} - \sqrt {{x^2} - x - 1} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {{x^2} + x - 1} \right) - \left( {{x^2} - x - 1} \right)}}{{\sqrt {{x^2} + x - 1} + \sqrt {{x^2} - x - 1} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x}}{{\left| x \right|\sqrt {1 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} + \left| x \right|.\sqrt {1 - \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x}}{{ - x\left( {\sqrt {1 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} + \sqrt {1 - \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} } \right)}}\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{2}{{ - \left( {\sqrt {1 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} + \sqrt {1 - \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} } \right)}}\\
= \dfrac{2}{{ - \left( {\sqrt 1 + \sqrt 1 } \right)}} = - 1
\end{array}\)
