Negative exponential growth Thanks for reading!
I feel like I have an okay enough understanding of $e$. 
This is how I understand it (wordy definition incoming, but I'm trying to make it as complete as possible):
By definition, $e$ is by how much some principal amount $P$ gets scaled by if we allow that principal to produce continuously, at a rate such that it produces itself over an interval of $\Delta t$, for an interval of $\Delta t$, while also allowing every piece produced in that interval to produce at that same rate (every piece produced in that interval would also produce itself if we allowed it to produce for $\Delta t$).
Idk if that made any sense, but...mathematically:
$$\lim_{n\rightarrow\infty} P(1+\frac{1}{n})^n=P*e$$
I can accept that by definition, that's what $e$ is.
However, what's befuddling me is, intuitively, WHY that would imply that:
$$\lim_{n\rightarrow\infty} P(1-\frac{1}{n})^n=P*\frac{1}{e}$$
Why is it that when something shrinks at a rate such that it would disappear in a single time interval $\Delta t$, but it does so continuously, after that time interval it'll have gotten divided by $e$?
Why not by some other number? 
Had I not known this, or seem some proof, I would've never been able to predict that $e$ would show up again. Yet, there's got to be an intuitive way to see that it would!
I understand that even though it's shrinking at a rate of "decaying by itself in a time interval" the principal won't have disappeared in the time interval, because just like with exponential growth where the "rate of production" is increasing, with exponential decay the "rate of decaying" of the principal is slowing down, as the rate of decay at any given time is equal to the total current amount, which is decreasing.
But, I can't see that the number that our principal would get divided by is $e$!!! 
After one time interval of $\Delta t$, why is it that when continuously shrinking at a rate equal to its current value, the principal shrinks by the same factor than it grew by in that same time interval with continuous growth?!?!
Why doesn't it shrink by some other factor? How can I intuitively see that $e$ would show up again?
Thanks!
 A: Let $a_n := (1+\frac 1n)^n$. We know this sequence converges to $e$.
It's intuitive that the sequence $b_n := (1-\frac 1n)^n$ converges to something between $0$ and $1$ inclusive, since the entire sequence lies between $0$ and $1$, so let's say it converges to $f$.
Then $a_nb_n = (1-\frac 1n)^n \cdot (1+\frac 1n)^n$ converges to $e\cdot f$.
But $a_nb_n = (1-\frac 1n)^n\cdot (1+\frac 1n)^n = ((1-\frac 1n)(1+\frac 1n))^n = (1-\frac 1{n^2})^n$. What does this converge to? Again, it needs to converge to something between $0$ and $1$, but this time we can use Bernoulli's inequality to find out that $(1-\frac 1{n^2})^n > 1 - \frac n{n^2} = 1-\frac 1n$. As $n$ grows large, this term goes to $1$, so by the squeeze theorem the sequence $a_nb_n$ has to converge to $1$.
Now we know that $e \cdot f = 1$, so $f = \frac 1e$.
A: $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.$$
So with $x=1$ you get $e^1=e$ and with $x=-1$ you get $e^{-1}=1/e$.
To see where the limit representation comes from, let's take the natural logarithm:
$$x=\lim_{n\to\infty}n\log\left(1+\frac{x}{n}\right).$$
Using the Taylor expansion for log, we have
$$\lim_{n\to\infty}n\left(\frac{x}{n}-\frac{x^2}{n^2}+\cdots\right)=\lim_{n\to\infty}x+O(1/n)=x.$$
Check out this link also.
A: Okay take the first case when calculating P*e. Now another thing you could do is instead of having $(1+\frac{1}{n})^n$ try instead $(1+\frac{1}{n})^{-n}$. As you might know anything raised to a negative exponent is just the reciprical of that. Then try to simplify that new expression via partial fraction decomposition.
