Đáp án:
$L=\lim\limits_{x\to 0}\left[\dfrac{(1+2018x)^{2019} -(1+2019x)^{2018}}{x^2}\right] = 2\,037\,171$
Giải thích các bước giải:
$\begin{array}{l}\quad L=\lim\limits_{x\to 0}\left[\dfrac{(1+2018x)^{2019} -(1+2019x)^{2018}}{x^2}\right]\\ \text{Áp dụng quy tắc L'Hôpital ta được:}\\ \quad L = \lim\limits_{x\to 0}\left[\dfrac{2019.2018(1+2018x)^{2018} -2018.2019(1+2019x)^{2017}}{2x}\right]\\ \to L = \dfrac{2019.2018}{2}\cdot \lim\limits_{x\to 0}\left[\dfrac{(1+2018x)^{2018} -(1+2019x)^{2017}}{x}\right]\\ \to L = \dfrac{2019.2018}{2}\cdot \lim\limits_{x\to 0}\left[\dfrac{2018.2018(1+2018x)^{2017} -2017.2019(1+2019x)^{2016}}{1}\right]\\ \to L = \dfrac{2019.2018}{2}\cdot \left[2018^2.(1+2018.0)^{2017} - 2017.2019(1 + 2019.0)^{2016} \right]\\ \to L = \dfrac{2019.2018}{2}\cdot 1\\ \to L= 2\,037\,171 \end{array}$
