mathop lim limitsx to pm giới hạn d2x^2 + 3x - 5 căn x^2 + 1 - 3x
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
Đáp án đúng: C
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2}\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{\sqrt {{x^2}} \sqrt {1 + \dfrac{1}{x}} - 3x}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2}\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{x\left( {\sqrt {1 + \dfrac{1}{x}} - 3} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}} \right)}}{{\sqrt {1 + \dfrac{1}{x}} - 3}}\end{array}\)
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: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( { - x} \right) = + \infty \\\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 + \dfrac{3}{x} - \dfrac{5}{{{x^2}}}}}{{\sqrt {1 + \dfrac{1}{x}} - 3}} = \dfrac{2}{{1 - 3}} = - 1 < 0\end{array} \right.\)\( \Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = - \infty \).
Vậy \(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = - \infty \).
Chọn C.
