\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