Solving for a variable inside a limit? I'm wondering how to solve for a variable inside a limit: 
My textbook defines $e$ implicitly as the number such that 
$$e\iff\displaystyle{\lim_{h\to 0}\frac{e^h-1}{h}}=1$$
I know from elsewhere $e$ can also be defined differently:
$$e=\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$$
I'm wondering how to solve for $e$ in the first definition. Can the second definition be derived from the first?
To solve for $e$ in the first definition, my initial direction would be to try and algebraically remove the limit on the left side of the equation. I can't see any way of doing this and don't have any other tricks up my sleeve. It would be enlightening if some other tricks are suggested or if my assumptions are told to be false. 
 A: Let us define the sequence
\begin{align}
a_n = \left(1+\frac{1}{n}\right)^n
\end{align}
then it's not hard to see that
\begin{align}
\left(1+\frac{1}{n}\right)^n\leq \left( 1+\frac{1}{n+1}\right)^{n+1}
\end{align}
but I will not do the calculation here. Moreover, the sequence it bounded since
\begin{align}
\left(1+\frac{1}{n} \right)^n =\sum^n_{k=0}\binom{n}{k}\frac{1}{n^k}=\sum^n_{k=0}\frac{1}{k!}\left( 1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right).
\end{align}
Using the fact that $2^{k-1}\leq k!$ for $k=2, 3, \ldots$ and
\begin{align}
\left( 1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right) \leq 1
\end{align}
 we see that
\begin{align}
\left(1+\frac{1}{n} \right)^n \leq \sum^n_{k=0} \frac{1}{2^{k-1}} \leq \sum^\infty_{k=0} \frac{1}{2^{k-1}}<\infty. 
\end{align}
Thus, it follows $a_n$ converges to some number $a$. 
Back to the problem. Since the limit
\begin{align}
\lim_{h\rightarrow 0} \frac{e^h-1}{h} = 1
\end{align}
holds, then
\begin{align}
\lim_{n\rightarrow \infty} n(e^{1/n}-1) = 1
\end{align}
where we replaced $h$ with $1/n$. Next, observe
\begin{align}
\lim_{n\rightarrow \infty} [n(e^{1/n}-1)-n(a_n^{1/n}-1)] = \lim_{n\rightarrow \infty}n(e^{1/n}-a_n^{1/n})=0
\end{align}
since
\begin{align}
n(a_n^{1/n}-1) = 1.
\end{align}
Finally, we have
\begin{align}
|e-a_n| = |e^{n/n}-a_n^{n/n}|= |e^{1/n}-a_n^{1/n}|\left|\sum^{n-1}_{k=0} e^{k/n}a_n^{(n-1-k)/n} \right|\leq n|e^{1/n}-a_n^{1/n}|\max\{e, a\}\rightarrow 0
\end{align}
as $n\rightarrow \infty$. Thus, $a_n \rightarrow e$. 
A: Ah I know exactly what you're saying and I had that trouble too. 
So we have: 
$$e\iff\displaystyle{\lim_{h\to 0}\frac{e^h-1}{h}}=1$$
$$e=\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$$
Let's examine the second one and do some manipulation:
$n = 1/h$
$$e=\lim_{h\to0}\left(1+h\right)^{1/h}$$
If we look at small values of $h$, 
$$e \approx (1+h)^{1/h}$$
Therefore,
$$e^h \approx 1+h$$
Replace this with $e^h$ in our first limit.
$$\lim_{h\to 0}\frac{e^h-1}{h} = \lim_{h\to 0}\frac{h+1-1}{h} = \lim_{h\to 0}\frac{h}{h} = 1$$
