Euler edits

Advanced/Euler's Number/parts/3 more e.tex

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+\section{More about $e$}
+
+\problem{}
+Show that
+$$
+	\lim_{n\to\infty}{
+		\bigg(
+			1 + \frac{1}{n+1}
+		\bigg)^n
+		= e
+	}
+$$
+
+\vfill
+
+
+
+
+\problem{}
+Show that
+$$
+	\lim_{n\to\infty}{
+		\bigg(
+			1 + \frac{1}{n}
+		\bigg)^{n+1}
+		= e
+	}
+$$
+
+\vfill
+
+
+
+
+\problem{}<inverse_e>
+Show that
+$$
+	\lim_{n\to\infty}{
+		\bigg(
+			1 - \frac{1}{n}
+		\bigg)^n
+		= \frac{1}{e}
+	}
+$$
+
+\begin{solution}
+	$
+		\lim_{n\to\infty}{(1 - \frac{1}{n})^n} =
+		\lim_{n\to\infty}{(\frac{n-1}{n})^{(-1)(-n)}}
+	$ \par
+
+	$
+	= \lim_{n\to\infty}{(\frac{n}{n- 1 })^{-n}}
+	= \lim_{n\to\infty}{(1 + \frac{1}{n-1})^{-n}}
+	$ \par
+
+	$
+	= \lim_{n\to\infty}{{(1 + \frac{1}{n-1})^{(n-1)(\frac{n}{n-1})}}^{-1}}
+	$ \par
+
+	$
+	= \frac{1}{e}
+	$
+\end{solution}
+
+\vfill
+
+
+
+
+\problem{}
+Show that
+$$
+	\lim_{n\to\infty}{
+		\bigg(
+			1 + \frac{x}{n}
+		\bigg)^n
+		= e^x
+	}
+$$
+Note that \ref{inverse_e} is a special case of this problem.
+
+\vfill
+\pagebreak
+
+
+
+\theorem{}
+The following important formula is proven in most calculus courses.
+
+$$e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}} = 2 + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
+
+\vfill
+
+
+
+
+\problem{}
+What are the first six digits of $e$?
+
+\begin{solution}
+	$e = 2.718\ 281\ 828$
+\end{solution}
+
+\vfill
+
+
+
+
+\definition{}
+If $f$ is a function, we say that $L$ is a limit of $f$ at $\infty$ if for every $\epsilon > 0$, we can find an $M \in \mathbb{R}$ so that $|f(x) - L| < \epsilon$ for $x > M$. \par
+If this is true, we say that  $L = \lim_{x\to\infty}{(f(x))}$.
+
+\vfill
+
+
+
+
+\problem{}
+Prove the following: \par
+Hint: If $x > 0$, then $\lfloor x \rfloor \leq x \leq \lceil x \rceil$
+
+$$ \lim_{x\to\infty}{\bigg(1 + \frac{1}{x}\bigg)^x} = e$$
+
+\vfill
+\pagebreak