Tìm giới hạn \(K=\lim\limits _{x \rightarrow+\infty}\left(\sqrt{x^{2}+1}+\sqrt{x^{2}-x}-2 x\right)\)
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Lời giải:
Báo sai\(\begin{array}{l} \\ \begin{array}{*{20}{l}} {{\rm{ Ta có : }}K = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 2{x^2} - x + 1 + 2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} - x} \right)} }}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x}}}\\ { = \mathop {\lim }\limits_{x \to + \infty } \frac{{4\left( {{x^4} - {x^3} + {x^2} - x} \right) - {{\left( {2{x^2} + x - 1} \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} - x} \right)} + 2{x^2} + x - 1} \right)}}}\\ { = \mathop {\lim }\limits_{x \to + \infty } \frac{{4\left( {{x^4} - {x^3} + {x^2} - x} \right) - {{\left( {2{x^2} + x - 1} \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} - x} \right)} + 2{x^2} + x - 1} \right)}}}\\ \begin{array}{l} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 8{x^3} + 7{x^2} - 2x - 1}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} - x} \right)} + 2{x^2} + x - 1} \right)}}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3}\left( { - 8 + \frac{7}{x} - \frac{2}{{{x^2}}} - \frac{1}{{{x^3}}}} \right)}}{{\left[ {x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + \sqrt {1 - \frac{1}{x}} + 2} \right)} \right]\left[ {{x^2}\left( {2\sqrt {1 + \frac{1}{{{x^2}}}} \sqrt {1 - \frac{1}{x}} + 2 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)} \right]}}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 8 + \frac{7}{x} - \frac{2}{{{x^2}}} - \frac{1}{{{x^3}}}}}{{\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + \sqrt {1 - \frac{1}{x}} + 2} \right)\left( {2\sqrt {1 + \frac{1}{{{x^2}}}} \sqrt {1 - \frac{1}{x}} + 2 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 8 + \frac{7}{x} - \frac{2}{{{x^2}}} - \frac{1}{{{x^3}}}}}{{2\left( {1 + \frac{1}{{{x^2}}}} \right)\sqrt {1 - \frac{1}{x}} + 2\sqrt {1 + \frac{1}{{{x^2}}}} + \frac{1}{x}\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{1}{{{x^2}}}\sqrt {1 + \frac{1}{{{x^2}}}} + 2\sqrt {1 + \frac{1}{{{x^2}}}} \left( {1 - \frac{1}{x}} \right) + 2\sqrt {1 - \frac{1}{x}} + \frac{1}{x}\sqrt {1 - \frac{1}{x}} - \frac{1}{{{x^2}}}\sqrt {1 - \frac{1}{x}} + 4\sqrt {1 + \frac{1}{{{x^2}}}} \sqrt {1 - \frac{1}{x}} + 4 + \frac{2}{x} - \frac{2}{{{x^2}}}}}\\ = \frac{{ - 8}}{{2 + 2 + 2 + 2 + 4 + 4}}\\ = - \frac{1}{2} \end{array} \end{array} \end{array}\)
