Tìm các giới hạn sau:

Giải chi tiết:

a) \(\mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1}  - 3x}}\)

\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1}  - 3x}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2}\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{\sqrt {{x^2}} \sqrt {1 + \dfrac{1}{x}}  - 3x}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2}\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{x\left( {\sqrt {1 + \dfrac{1}{x}}  - 3} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{x\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{\sqrt {1 + \dfrac{1}{x}}  - 3}}\end{array}\)

Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty \\\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}}}{{\sqrt {1 + \dfrac{1}{x}}  - 3}} = \dfrac{2}{{1 - 3}} =  - 1 < 0\end{array} \right.\)\( \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1}  - 3x}} =  - \infty \).

\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1}  - 3x}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{{x^2}\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{\sqrt {{x^2}} \sqrt {1 + \dfrac{1}{x}}  - 3x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{{x^2}\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{ - x\left( {\sqrt {1 + \dfrac{1}{x}}  + 3} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - x\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{\sqrt {1 + \dfrac{1}{x}}  + 3}}\end{array}\)

Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( { - x} \right) =  + \infty \\\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}}}{{\sqrt {1 + \dfrac{1}{x}}  + 3}} = \dfrac{2}{{1 + 3}} = \dfrac{1}{2} > 0\end{array} \right.\)\( \Rightarrow \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1}  - 3x}} =  + \infty \).

Vậy \(\mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1}  - 3x}} =  \mp \infty \).

Giải chi tiết:

\(\mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}}\)

\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {{x^6}} \sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{{x^3}}}.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^3}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{x.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\end{array}\)

Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } {x^2} =  + \infty \\\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = 1 > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} =  + \infty \)

\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {{x^6}} \sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{{x^3}}}.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - {x^3}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{x.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - {x^2}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\end{array}\)

Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( { - {x^2}} \right) =  - \infty \\\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = 1 > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} =  - \infty \).

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

Giải chi tiết:

\(\mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{{2x\sqrt {{x^2} + x}  - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}}\)

\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2x\sqrt {{x^2} + x}  - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2{x^2}\sqrt {1 + \dfrac{1}{x}}  - x\sqrt {9 + \dfrac{1}{{{x^2}}}} }}{{{x^2}\left( {2 + \dfrac{1}{x}} \right)\left( {1 - \dfrac{2}{x}} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2\sqrt {1 + \dfrac{1}{x}}  - \dfrac{1}{x}\sqrt {9 + \dfrac{1}{{{x^2}}}} }}{{\left( {2 + \dfrac{1}{x}} \right)\left( {1 - \dfrac{2}{x}} \right)}} = \dfrac{{2 - 0}}{{2.1}} = 1\end{array}\)

\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{2x\sqrt {{x^2} + x}  - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - 2{x^2}\sqrt {1 + \dfrac{1}{x}}  + x\sqrt {9 + \dfrac{1}{{{x^2}}}} }}{{{x^2}\left( {2 + \dfrac{1}{x}} \right)\left( {1 - \dfrac{2}{x}} \right)}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - 2\sqrt {1 + \dfrac{1}{x}}  + \dfrac{1}{x}\sqrt {9 + \dfrac{1}{{{x^2}}}} }}{{\left( {2 + \dfrac{1}{x}} \right)\left( {1 - \dfrac{2}{x}} \right)}} = \dfrac{{ - 2 + 0}}{{2.1}} =  - 1\end{array}\)

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