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\title{Continuity}
\date{September 8, 2021}

\begin{document}
\ \vspace{-.25in}

\maketitle

\vspace{-.15in}

\dfn{
A function $f$ is \textbf{continuous} at $c$ if $\displaystyle \boxed{\lim_{x \to c} f(x) = f(c)}.$
This means that $f$ is only continuous at $c$ if all 3 of the following are true:
\begin{enumerate}
\item $f(c)$ is defined,
\item $\lim_{x \to c} f(x)$ exists, and
\item $\lim_{x \to c} f(x) = f(c)$.
\end{enumerate}
}

\prob{
Explain (with reference to the list above) why each of the following functions is \textbf{not} continuous at $c = 2$.
\begin{enumerate}[a)]
\item $\displaystyle f(x) = \frac{1}{2-3x+x^2}$

\vfill

%\item[b) ] $\displaystyle f(x) = \begin{cases} \frac{1}{2-3x+x^2} & \text{ if } x \neq 2 \\ 1 & \text{ if } x = 2\end{cases}$

%\vfill

\item $\displaystyle g(x) = \frac{2-x}{2-3x+x^2}$

\vfill

\item $\displaystyle h(x) = \begin{cases} \frac{2-x}{2-3x+x^2} & \text{ if } x \neq 2 \\ 1 & \text{ if } x = 2\end{cases}$

\vfill

\end{enumerate}
}

\newpage

\prob{
What value $c$ would make the function continuous at $0$?
\begin{enumerate}[a)]

\item $\displaystyle f(x) = 
\begin{cases} 
c - x^2 & \text{ if } x \geq 0 \\ 
\cos x & \text{ if } x < 0 
\end{cases} $

\vfill

\item  $\displaystyle g(x) = 
\begin{cases} 
\frac{\sqrt{4+x} - 2}{x} & \text{ if } x \neq 0 \\ 
c & \text{ if } x = 0 
\end{cases} $

\vfill

\end{enumerate}
}

\prob{
Find the points at which the function
%{\sin\left( \frac{1}{\pi x} \right) }$ 
is \textbf{not} continuous.

\begin{enumerate}[a)]

\item $\displaystyle f(x) = \begin{cases} 
x &\text{ if } x<-1 \\
x^3 &\text{ if } -1 \leq x < 1 \\
\frac{1}{x} &\text{ if } 1 \leq x
\end{cases} $

\vfill

\item $\displaystyle g(x) = \frac{1}{1-\cos x}$

\vfill

\item How does your answer to part b change if we restrict the domain of $g$ to the interval $(0,\pi)$?

\end{enumerate}
}

\vspace{1in}

\end{document}
