mathop lim limitsx to + giới hạn dx^2020 + 2021x + 12x^2020 - 2022x + 2
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^{2020}} + 2021x + 1}}{{2{x^{2020}} - 2022x + 2}}\)
Đáp án đúng: D
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^{2020}} + 2021x + 1}}{{2{x^{2020}} - 2022x + 2}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^{2020}}\left( {1 + \dfrac{{2021}}{{{x^{2019}}}} + \dfrac{1}{{{x^{2020}}}}} \right)}}{{{x^{2020}}\left( {2 - \dfrac{{2022}}{{{x^{2019}}}} + \dfrac{2}{{{x^{2020}}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 + \dfrac{{2021}}{{{x^{2019}}}} + \dfrac{1}{{{x^{2020}}}}}}{{2 - \dfrac{{2022}}{{{x^{2019}}}} + \dfrac{2}{{{x^{2020}}}}}} = \dfrac{1}{2}\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^{2020}} + 2021x + 1}}{{2{x^{2020}} - 2022x + 2}} = \dfrac{1}{2}\).
Chọn D.
