Đáp án:
$\lim\limits_{x\to 1}\left(\dfrac{2017}{1 - x^{2017}} - \dfrac{2018}{1-x^{2018}}\right) = -\dfrac12$
Giải thích các bước giải:
$\quad \lim\limits_{x\to 1}\left(\dfrac{2017}{1 - x^{2017}} - \dfrac{2018}{1-x^{2018}}\right)$
$= \lim\limits_{x\to 1}\left[\dfrac{2017}{(1 - x)(x^{2016}+ x^{2015} + \dots + x +1)} - \dfrac{2018}{(1-x)(x^{2017}+ x^{2016} + \dots + x +1)}\right]$
$= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots +1)}{(1-x)(x^{2016} +\dots + 1)(x^{2017} + \dots + 1)}$
$= \lim\limits_{x\to 1}\dfrac{1}{(x^{2016} +\dots + 1)(x^{2017} + \dots + 1)}\cdot \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - x -1}{1-x}$
$= \dfrac{1}{2017.2018}\cdot \lim\limits_{x\to 1}\dfrac{2017.2017x - 2016x^{2015} - \dots - 2x -1}{-1}\qquad (l'Hôpital)$
$= \dfrac{1}{2017.2018}\cdot ( 2016 +\dots + 2 +1- 2017.2017)$
$= \dfrac{1}{2017.2018}\cdot\left(\dfrac{2016.2017}{2} - 2017.2017\right)$
$=\dfrac{1}{2017.2018}\cdot (-2017.1009)$
$= -\dfrac12$
