m3.tex

\documentclass{article}
\usepackage{amsmath}
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\title{MATH 4338 Main Problem 3}
\date{}
\author{Andy Lu}

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\usepackage{amsthm}
\newtheorem{theorem}{Question}

\begin{document}
  \maketitle
  \begin{theorem}
    Suppose $f$ and $h$ are continuous at $a$ and $f(x) \leq g(x) \leq h(x)$ for
    $|x-a|<k$. If $f(a) = h(a)$, show that $g$ is continuous at a.
  \end{theorem}
  
  \begin{proof}
    Assume $f, h$ are continuous at $a$. Assume $f(x) \leq g(x) \leq h(x)$ for 
    $|x-a| < k$. Lastly, assume $f(a) = h(a)$. \\
    
    By the definition of continuity,
    \begin{align*}
      & & \lim_{x \rightarrow a} f(x) = f(a) & 
      & \lim_{x \rightarrow a} h(x) = h(a) & &
    \end{align*}
    Then by the Sandwich Theorem, $\lim_{x \rightarrow a} g(x) = g(a)$.
    Pick $x = a$ for $|x-a| < k$. As this inequality holds for $x=a$,
    $f(a) = g(a) = h(a)$. Furthermore, this means $g(a)$ exists. By the limit 
    definition of continuity, since $g(a)$ exists and 
    $$\lim_{x \rightarrow a} g(x) = g(a)$$ $g(x)$ is continuous at a.
  \end{proof}
\end{document}