tính giới hạn lim 0,1+0,11+0,111+….+0,11…….11 / 2n+1 09/11/2021 Bởi Aaliyah tính giới hạn lim 0,1+0,11+0,111+….+0,11…….11 / 2n+1
Đáp án: $+\infty$ Giải thích các bước giải: Ta có: $\lim 0.1+0.11+0.111+…+0.11\cdots11$ $=\lim \dfrac19(0.9+0.99+0.999+…+0.99\cdots99)$ $=\lim \dfrac19(1-0.1+1-0.01+1-0. 001+…+1-0. 00\cdots 01)$ $=\lim \dfrac19((2n+1)-(0.1+0.01+0. 001+…+0. 00\cdots 01))$ $=\lim \dfrac19((2n+1)-(10^{-1}+10^{-2}+10^{-3}+…+10^{-(2n+1)}))$ $=\lim \dfrac19((2n+1)-\dfrac{1-10^{-(2n+1)}}{10-1})$ $=\lim \dfrac19((2n+1)-\dfrac{1-10^{-(2n+1)}}{9})$ $=\lim \dfrac{9(2n+1)-1+10^{-(2n+1)}}{81}$ $= \dfrac{+\infty-1+0}{81}$ $=+\infty$ Bình luận
Đáp án: $+\infty$
Giải thích các bước giải:
Ta có:
$\lim 0.1+0.11+0.111+…+0.11\cdots11$
$=\lim \dfrac19(0.9+0.99+0.999+…+0.99\cdots99)$
$=\lim \dfrac19(1-0.1+1-0.01+1-0. 001+…+1-0. 00\cdots 01)$
$=\lim \dfrac19((2n+1)-(0.1+0.01+0. 001+…+0. 00\cdots 01))$
$=\lim \dfrac19((2n+1)-(10^{-1}+10^{-2}+10^{-3}+…+10^{-(2n+1)}))$
$=\lim \dfrac19((2n+1)-\dfrac{1-10^{-(2n+1)}}{10-1})$
$=\lim \dfrac19((2n+1)-\dfrac{1-10^{-(2n+1)}}{9})$
$=\lim \dfrac{9(2n+1)-1+10^{-(2n+1)}}{81}$
$= \dfrac{+\infty-1+0}{81}$
$=+\infty$