From b8711ffb6a3890ae83737051093d4a7958c61c57 Mon Sep 17 00:00:00 2001 From: Mark Date: Thu, 12 Jan 2023 09:41:51 -0800 Subject: [PATCH] Fixed old boxlinehack command --- Advanced/Euler's Number/main.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/Advanced/Euler's Number/main.tex b/Advanced/Euler's Number/main.tex index d2bb369..bdc0266 100755 --- a/Advanced/Euler's Number/main.tex +++ b/Advanced/Euler's Number/main.tex @@ -115,7 +115,7 @@ 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 $\lim_{n\to\infty}a_n = 0$ - \boxlinehack + \linehack{} $\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \\ @@ -149,7 +149,7 @@ 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.