Fixed old boxlinehack command

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.