Tính giới hạn của các hàm số sau: a) \(\mathop {\lim }\limits_{x \to + \infty }

Câu hỏi số 648800:

Vận dụng

Tính giới hạn của các hàm số sau:

a) \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3} + 3x + 1}}{{ - 6{x^3} - 6{x^2} + 2}}\)

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

c) \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} - 5}}{{3{x^5} - 4{x^2} + x + 1}}\)

d) \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{(2x - 3)}^{10}} \cdot {{(4x + 1)}^{12}}}}{{{{(x - 5)}^{22}}}}\)

e) \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{5x + 1}}{{3x + \sqrt {4{x^2} + 1} }}\)

f) \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{5x + 1}}{{4x + \sqrt {4{x^2} + 1} }}\)

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Câu hỏi:648800

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Giải chi tiết

a) \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3} + 3x + 1}}{{ - 6{x^3} - 6{x^2} + 2}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3}\left( {1 + \dfrac{3}{{{x^2}}} + \dfrac{1}{{{x^3}}}} \right)}}{{{x^3}\left( { - 6 - \dfrac{6}{x} + \dfrac{2}{{{x^3}}}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 + \dfrac{3}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}{{ - 6 - \dfrac{6}{x} + \dfrac{2}{{{x^3}}}}} = \dfrac{{ - 1}}{6}\)

b) \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3} - 2{x^2} + 3x - 5}}{{2{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3}\left( {1 - \dfrac{2}{x} + \dfrac{3}{{{x^2}}} - \dfrac{5}{{{x^3}}}} \right)}}{{{x^2}\left( {2 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {1 - \dfrac{2}{x} + \dfrac{3}{{{x^2}}} - \dfrac{5}{{{x^3}}}} \right)}}{{\left( {2 + \dfrac{1}{{{x^2}}}} \right)}}\)

\( = \mathop {\lim }\limits_{x \to + \infty } x \cdot \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 - \dfrac{2}{x} + \dfrac{3}{{{x^2}}} - \dfrac{5}{{{x^3}}}}}{{2 + \dfrac{1}{{{x^2}}}}} = + \infty \)

c) \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} - 5}}{{3{x^5} - 4{x^2} + x + 1}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^2}\left( {2 - \dfrac{5}{{{x^2}}}} \right)}}{{{x^5}\left( {3 - \dfrac{4}{{{x^3}}} + \dfrac{1}{{{x^4}}} + \dfrac{1}{{{x^5}}}} \right)}}\)\( = \mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{{x^3}}} \cdot \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 - \dfrac{5}{{{x^2}}}}}{{3 - \dfrac{4}{{{x^3}}} + \dfrac{1}{{{x^4}}} + \dfrac{1}{{{x^5}}}}} = 0 \cdot \dfrac{2}{3} = 0.\)

d) \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{(2x - 3)}^{10}}{{(4x + 1)}^{12}}}}{{{{(x - 5)}^{22}}}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^{10}}{{\left( {2 - \dfrac{3}{x}} \right)}^{10}}{x^{12}}{{\left( {4 + \dfrac{1}{{{x^{12}}}}} \right)}^{12}}}}{{{x^{22}}{{\left( {1 - \dfrac{5}{{{x^{22}}}}} \right)}^{22}}}}\)

\( = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{{\left( {2 - \dfrac{3}{x}} \right)}^{10}}{{\left( {4 + \dfrac{1}{{{x^{12}}}}} \right)}^{12}}}}{{{{\left( {1 - \dfrac{5}{{{x^{22}}}}} \right)}^{22}}}} = \dfrac{{{{(2 - 0)}^{10}}{{(4 + 0)}^{12}}}}{{{{(1 - 0)}^{22}}}} = {2^{34}}\)

e)\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{5x + 1}}{{3x + \sqrt {4{x^2} + 1} }} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{5x + 1}}{{3x + |x|\sqrt {4 + \dfrac{1}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{5x + 1}}{{3x + x\sqrt {4 + \dfrac{1}{{{x^2}}}} }}\)

\( = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {5 + \dfrac{1}{x}} \right)}}{{x\left( {3 + \sqrt {4 + \dfrac{1}{{{x^2}}}} } \right)}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{5 + \dfrac{1}{x}}}{{3 + \sqrt {4 + \dfrac{1}{{{x^2}}}} }} = \dfrac{{5 + 0}}{{3 + \sqrt {4 + 0} }} = 1\).

f) \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt[3]{{8{x^3} - 5{x^2} + 1}}}}{{4x + \sqrt {4{x^2} + 1} }} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x \cdot \sqrt[3]{{8 - \dfrac{5}{x} + \dfrac{1}{{{x^3}}}}}}}{{4x + |x|\sqrt {4 + \dfrac{1}{x}} }} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\sqrt[3]{{8 - \dfrac{5}{x} + \dfrac{1}{{{x^3}}}}}}}{{4x - x\sqrt {4 + \dfrac{1}{x}} }}\)

\( = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt[3]{{8 - \dfrac{5}{x} + \dfrac{1}{{{x^3}}}}}}}{{4 - \sqrt {4 + \dfrac{1}{x}} }} = \dfrac{{\sqrt[3]{{8 - 0 + 0}}}}{{4 + \sqrt {4 + 0} }} = \dfrac{2}{6} = \dfrac{1}{3}.\)

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