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\centerline{MATH 133 -- Calculus 1}
\centerline{Limits Numerically and Graphically}
\centerline{September 18, 2019}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
The tables of values of average velocities
that we were looking at last time 
are aimed at trying to estimate values of instantaneous velocities: 
\begin{equation}
v_{inst} = \lim_{\Delta t\to 0} \frac{\Delta x}{\Delta t} = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t},\ 
{\rm or}\label{instvel},
\end{equation}
and slopes of tangents:
\begin{equation}
m_{tan} = \lim_{\Delta x\to 0} \frac{f(a + \Delta x) - f(a)}{\Delta x}.\label{mtan}
\end{equation}
For the next few days we will be looking at this process of estimating limits. Eventually
we will develop methods to compute limits exactly in many cases.  To focus on the limiting
process, we will momentarily take a step back from the particular limits in \eqref{instvel} and \eqref{mtan}
and look at limits
\begin{equation}
\label{genlim}
\lim_{x\to c} f(x)
\end{equation}
in general.  

To start we will look at numerical and graphical methods to understand these general limits.
Recall the general idea from our video for today -- to say in \eqref{genlim} that
$\lim_{x\to c} f(x) = L$, we want to be convinced that $|f(x) - L|$ can be made
arbitrarily small by taking $x$ sufficiently close to $c$, {\it but we won't just substitute
$x = c$ because we want to allow for the limit to exist even when $f(c)$ does not exist.}
(This is the situation for instance in \eqref{instvel} if we look at $\frac{x(t + \Delta t) - x(t)}{\Delta t}$
as a function of $\Delta t$ -- we want to all for a limit as $\Delta t \to 0$ to exist even though
direct subsitution gives us $0/0$, which is undefined(!)  Similarly for \eqref{mtan}, where the
$\frac{f(a+\Delta x) - f(a)}{\Delta x}$ is viewed as a function of $\Delta x$.)  
\vskip 10pt
\noindent
{\it Questions}
\vskip 10pt
\begin{enumerate}
\item[(1)] Complete the following table and use your results to guess the limit 
$\lim_{x\to 4} \frac{x^2 - 16}{x - 4}$.  Use at least 5 decimal places in all calculations for these.
$$\begin{array}{cccccccccccccc}
\hline
x                     &\vline& 4.1   &\vline& 4.01 &\vline& 4.001 &\vline& 3.999 &\vline& 3.99 &\vline& 3.9 \\ \hline
\frac{x^2 - 16}{x - 4}&\vline&         &\vline&          &\vline&           &\vline&      &\vline&  &\vline& \\ \hline 
\end{array}$$
\item[(2)]   Complete the following table and use your results to guess the limit $\lim_{x\to 0} \frac{\cos(x) - 1}{x}$
Use at least 5 decimal places in all calculations for these.
$$\begin{array}{cccccccccccccc}
\hline
x                   &\vline& 0.1 &\vline& 0.01 &\vline& 0.001 &\vline& -0.001 &\vline& -0.01 &\vline& -0.1\\ \hline
\frac{\cos(x)-1}{x} &\vline&     &\vline&      &\vline&        &\vline&     &\vline&       &\vline&       \\ \hline 
\end{array}$$
\item[(3)]  The limit in question (2) can be interpreted as a limit of the form from \eqref{mtan} before.  
What is the function $f(x)$ there?  (Hint: It's not $\frac{\cos(x) - 1}{x}$.)  Does your guess of the value
of the limit agree with what you would expect from the graph of $y = f(x)$ for your $f$?) 
\item[(4)]  Refer to the graph above.  What are the limits 
$\lim_{x\to 1^+} f(x)$, $\lim_{x\to 1^-} f(x)$, $\lim_{x \to 2^+} f(x)$, and $\lim_{x\to 2^-} f(x)$.  Does
either $\lim_{x\to 1} f(x)$ or $\lim_{x\to 2} f(x)$ exist?  Why or why not?
\begin{figure}
\begin{center}
\includegraphics[width=.4\linewidth]{Discont.png}
\end{center}
\caption{Figure for Question 4}
\end{figure}

\end{enumerate}
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