\documentclass{amsbook} \newtheorem{theorem}{Theorem}[chapter] \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{section}{chapter} \numberwithin{equation}{chapter} \begin{document} % \begin{equation*} % \begin{split} % \end{split} % \end{equation*} \frontmatter \title{Minqi's Solutions to the Problems of the National Postgraduate Entrance Examination of China} \author{Minqi Pan\footnote{Website: http://www.minqi-pan.com}} \date{\today} \maketitle \setcounter{page}{4} \tableofcontents \mainmatter \chapter{Mathematics II, 2018} \section{Problem 1} Discover a possible pair of $a$ and $b$ that makes $\lim_{x\to 0}\left(e^x+ax^2+bx\right)^{\frac{1}{x^2}}=1.$ \begin{proof} We claim that$a=-\frac{1}{2},b=-1.$ is such a possible pair. By the definition of the exponential function $e^x=\sum_{n=0}^\infty \frac{x^n}{n!},$ we have$e^x-\frac{1}{2}x^2-x=1+\sum_{n=3}^\infty \frac{x^n}{n!}.$ Define $g(x)=\sum_{n=3}^\infty \frac{x^n}{n!}.$ Then $\left(e^x-\frac{1}{2}x^2-x\right)^{\frac{1}{x^2}}=\left[\left(1+g(x)\right)^\frac{1}{g(x)}\right]^\frac{g(x)}{x^2}.$ By the definition of the number $e$$e=\lim_{x\to 0}(1+x)^\frac{1}{x}$ and the fact that $\lim_{x\to 0}g(x)=0,$ we have $\lim_{x\to 0}\left(1+g(x)\right)^\frac{1}{g(x)}=e.$ Finally, since $\lim_{x\to 0}\frac{g(x)}{x^2}=0,$ we conclude that $\lim_{x\to 0}\left[\left(1+g(x)\right)^\frac{1}{g(x)}\right]^\frac{g(x)}{x^2}=e^0=1.$ \end{proof} \end{document}