Text Formatting ⇒ Newbie Question Here

[JohnScotch]

Hi,

I am using LATEX for just the second time, as part of my Real Analysis course.

I thought I found the correct instructions on how to set this up (it is a very simple proof after having just gotten the definition of continuity), but I get a bunch of "Missing $ inserted" errors and one "Bad math environment delimiter" error.

Code: Select all

 \documentclass[12pt]{amsart}

\usepackage{amsmath}

\begin{document}

\title{M4 Written Assignment Exercise 1}

\author{John J. Scotch II}
\address{SUNY Empire State College\\
Center for Distance Learning\\
Saratoga Springs, New York}
\email{jjscotch2@gmail.com}

\date{March 27, 2022}

\maketitle


Use definition 4.2.1 to supply a proof for the following limit statement: $\lim_{x}^{1} 3x + 4$.

Let $\epsilon > 0$. We need to select a \delta such that $0 < |x - 1| < \delta$ implies that $|f(x) - 7| < \epsilon$.

We note that

\begin{equation}
\|f(x) - 7| = |(3x + 4) - 7| = |3x - 3| = 3|x - 1|.
\end{equation}

Thus, if we choose $\delta = \( \frac{\epsilon}{3} \)$, then $0 < |x - 1| < \delta$ implies that $|f(x) - 7| = 3|x - 1| < 3(\frac{\epsilon}{3} \) = \epsilon$. \qed

Can anyone help me figure out why I am getting errors?

Thanks, J.J.

[Bartman]

Please build a minimal working example and use the code tags to mark the code so one can test it right here in the forum.

The commands for greek letters (here \delta) used in the example must be in math mode.

A hint for the last line: You can enter inline math mode with either $ or \(.

A suggestion:

Code: Select all

\documentclass[12pt]{amsart}

%\usepackage{amsmath}% is already loaded by the class

\title{M4 Written Assignment Exercise 1}

\author{John J. Scotch II}
\address{SUNY Empire State College\\
Center for Distance Learning\\
Saratoga Springs, New York}
\email{jjscotch2@gmail.com}

\date{March 27, 2022}

\begin{document}
\maketitle
Use definition 4.2.1 to supply a proof for the following limit statement: 
$\lim_{x}^{1} 3x + 4$.

Let $\epsilon > 0$. We need to select a $\delta$ such that $0 < |x - 1| < \delta$ 
implies that $|f(x) - 7| < \epsilon$.

We note that

\begin{equation}
\| f(x) - 7| = |(3x + 4) - 7| = |3x - 3| = 3|x - 1|.
\end{equation}
Thus, if we choose $\delta = ( \frac{\epsilon}{3} )$, then $0 < |x - 1| < \delta$ 
implies that $|f(x) - 7| = 3|x - 1| < 3(\frac{\epsilon}{3} ) = \epsilon$. \qed
\end{document}

[JohnScotch]

Bartman, thank you so much!

I did not know about the minimal working examples (and I get sent to a page that can't be reached when I click on it), but I will make sure I figure that out before posting again.

This is perfect, and I can compare what you did to what I had and figure out where I went wrong, although I will save that for tomorrow when my head isn't pounding.

Thanks again!

- J.J.