Why does this approximation work (and why does it fail)? I have a function
$$f(x)=\frac{e^{-x}}{x}$$
and I am trying to find an expression for the inverse function $f^{-1}(x)$.
So far I have come up with the approximation:
$$\hat{f}^{-1}(x)=\left( \left( x+v \right )^e-v^e \right)^{-\frac{1}{e}}$$
where $v \approx 0.83$
This formula is a good approximation for $x>0$. I basically came up with this formula from tinkering, but I'm not exactly sure what the relation is between the e-th root of an expression and e to the power of said expression. Does it have anything to do with the limit definition of e:
$$ \lim_{n \rightarrow \infty}{\left (1+\frac{1}{n} \right)^n} $$
However, this is just an approximation — an approximation that fails. Different values of $v$ offer better approximations, and I think that $v$ is related to $e$ in some way, however I'm not sure what way that is. Why does my function come so close, but also fail? Is this just a special case?
Here is a diagram showing $f^{-1}(x)$ (red) and $\hat{f}^{-1}(x)$ (blue).

 A: My opinion is that you just happened to stumble upon a function which is asymptotically equivalent to the actual inverse. 
The actual inverse to your equation is given by the Lambert W function. The principal branch of the Lambert W has a Taylor series expansion near $0$ given by
$$W(x) = x-x^2+\mathcal{O}(x^3).$$
The actual inverse to your function $f(x)$ is given by
$$f^{-1}(x) = W\left(\frac{1}{x}\right),$$
so for large $x$, we have
$$f^{-1}(x) = \frac{1}{x} - \frac{1}{x^2} +\mathcal{O}\left(\frac{1}{x^3}\right).$$
Your approximate inverse just happened to go as 
$$\hat{f}^{-1}(x) \sim \frac{1}{x+v} \sim \frac{1}{x} - \frac{v}{x^2} + \mathcal{O}\left(\frac{v^2}{x^3}\right),$$ 
where $v$ happens to be relatively close to $1$, which is the main reason why the approximation seems nice. 
In fact, if you were to plot the relative error between $f^{-1}$ and $1/x-1/x^2$ versus the relative error between $\hat{f}^{-1}$ and $f^{-1}$, you will find that the former is a much better approximation for $x \gtrsim 10$, as it must be since it is the Taylor polynomial. It is not surprising to me that you can find a better approximation for a small segment of the function by varying the value of $v$. Your approximation is best around $x\approx 2.3665$, which doesn't seem to be any special value for $W$, so I am inclined to say that you were rather lucky in finding a good approximation, in conjunction with the look-elsewhere effect. As you've admitted yourself, differing values of $v$ will give differing degrees of approximation, and I don't think there is anything particularly special going on. If you could tell us how you obtained this approximation, we might be able to say something more concrete.
Incidentally, if you also want an approximation which works for $x \ll 1$, then you might try the asymptotic expansion
$$W(1/x) \sim -\ln(-x\ln x)-\frac{\ln\left(-\ln x\right)}{\ln x} + \mathcal{O}\left(\frac{\ln\left(-\ln x\right)}{(\ln x)^2}\right).$$
