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Fixed old boxlinehack command

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Mark 2023-01-12 09:41:51 -08:00
parent 9e99560b38
commit b8711ffb6a
1 changed files with 2 additions and 2 deletions
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4
Advanced/Euler's Number/main.tex
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@ -115,7 +115,7 @@
We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \\ We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \\
It is now clear that $\lim_{n\to\infty}a_n = 0$ It is now clear that $\lim_{n\to\infty}a_n = 0$
\boxlinehack \linehack{}
$\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \\ $\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \\
@ -149,7 +149,7 @@
Therefore, if both $A$ and $B$ are limits of $[a_n]$, $A$ and $B$ must be equal. Therefore, if both $A$ and $B$ are limits of $[a_n]$, $A$ and $B$ must be equal.
\boxlinehack \linehack{}
\textsuperscript{*}Note that we can also set $\epsilon = \frac{|B-A|}{2}$, which gives $|B-A| < |B-A|$. This is also false, and the proof still works. However, the extra $\frac{\hspace{2em}}{2}$ gives a clearer explanation. \textsuperscript{*}Note that we can also set $\epsilon = \frac{|B-A|}{2}$, which gives $|B-A| < |B-A|$. This is also false, and the proof still works. However, the extra $\frac{\hspace{2em}}{2}$ gives a clearer explanation.