mathop lim limitsx to 2 x^2020 - 2^2020x^2021 - 2^2021
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^{2020}} - {2^{2020}}}}{{{x^{2021}} - {2^{2021}}}}\)
Đáp án đúng: A
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^{2020}} - {2^{2020}}}}{{{x^{2021}} - {2^{2021}}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {{x^{2019}} + {x^{2018}}.2 + {x^{2017}}{{.2}^2} + ... + x{{.2}^{2018}} + {2^{2019}}} \right)}}{{\left( {x - 2} \right)\left( {{x^{2020}} + {x^{2019}}.2 + {x^{2018}}{{.2}^2} + ... + x{{.2}^{2019}} + {2^{2020}}} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^{2019}} + {x^{2018}}.2 + {x^{2017}}{{.2}^2} + ... + x{{.2}^{2018}} + {2^{2019}}}}{{{x^{2020}} + {x^{2019}}.2 + {x^{2018}}{{.2}^2} + ... + x{{.2}^{2019}} + {2^{2020}}}}\\ = \dfrac{{{2^{2019}} + {2^{2018}}.2 + {2^{2017}}{{.2}^2} + ... + {{2.2}^{2018}} + {2^{2019}}}}{{{2^{2020}} + {2^{2019}}.2 + {2^{2018}}{{.2}^2} + ... + {{2.2}^{2019}} + {2^{2020}}}}\\ = \dfrac{{{2^{2019}}.2020}}{{{2^{2020}}.2021}} = \dfrac{{1010}}{{2021}}\end{array}\)
Chọn A.
