Ta có A=(x2013−1)−2(x−1)(x2014−1)−2(x−1)=(x−1)[x2012+x2011+...+1−2](x−1)[x2013+x2012+...+1−2]Vậy $\lim \limits_{x\to1}A=\lim \limits_{x\to1} \dfrac{x^{2012} +x^{2011}+...+1-2}{x^{2013} +x^{2012}+...+1-2}=\dfrac{2012-2}{2013-2}=\dfrac{2010}{2011}$
Ta có
A=(x2013−1)−2(x−1)(x2014−1)−2(x−1)=(x−1)[x2012+x2011+...+1−2](x−1)[x2013+x2012+...+1−2]Vậy $\lim \limits_{x\to1}A=\lim \limits_{x\to1} \dfrac{x^{2012} +x^{2011}+...+1-2}{x^{2013} +x^{2012}+...+1-2}=\dfrac{2012-2}{2013-2}=\dfrac{201
1}{201
2}$