$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x} = e,\ $ and $\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{x} = \frac{1}{e}.\ $ Is this generalisable? 
Suppose $f(x)\to\infty,\ g(x)\to\infty,\ $ both as $\ x\to\infty,\ $
and $\lim_{x\to\infty}\left(1+\frac{1}{f(x)}\right)^{g(x)} = c>1.\ $
Then does $\lim_{x\to\infty}\left(1-\frac{1}{f(x)}\right)^{g(x)} = \frac{1}{c}\ ?$

The proof I know/like for $\ f(x) = x\ $ and $\ g(x)=x\ $ is as follows:
Suppose $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x} = e.\ $ Then
\begin{align} \lim_{x\to\infty}\frac{1}{\left(1-\frac{1}{x}\right)^x}=\lim_{x\to\infty}\left(\frac{x}{x-1}\right)^x=\lim_{(x+1)\to\infty}\left(\frac{x+1}{x}\right)^{x+1}\\
\\
=\lim_{(x+1)\to\infty}\left(\left(\frac{x+1}{x}\right)^x\ \cdot\ \left(\frac{x+1}{x}\right)\right)=\lim_{x\to\infty}\left(\left(1+\frac{1}{x}\right)^x\ \cdot\ \left(1+\frac{1}{x}\right)\right) = e\cdot1 = e\\
\end{align}
and we are done.
I tried the same method with the generalisation:
Suppose $\lim_{x\to\infty}\left(1+\frac{1}{f(x)}\right)^{g(x)} = c>1.\ $ Then
$$ \lim_{x\to\infty}\frac{1}{\left(1-\frac{1}{f(x)}\right)^{g(x)}}=\lim_{x\to\infty}\left(\frac{f(x)}{f(x)-1}\right)^{g(x)}=\lim_{(x+1)\to\infty}\left(\frac{f(x+1)}{f(x+1)-1}\right)^{g(x+1)}=\ ?$$
Maybe this works with some Binomial or Taylor expansion, but I doubt it? Also I'm not even sure if the result it true...
 A: Suppose that $f(x),g(x) \to \infty$ as $x \to \infty$ , and $$
\lim_{x \to \infty} \left(1 + \frac 1{f(x)}\right)^{g(x)} = c>1
$$
where we note that the term is well defined since $f(x),g(x)$ are positive for large enough $x$. We want to prove that $\lim_{x \to \infty} \left(1 - \frac 1{f(x)}\right)^{g(x)} = \frac 1c$. The key relation is :
$$
\left(1 + \frac 1{f(x)}\right)^{g(x)} = \left(\left(1 + \frac 1{f(x)}\right)^{f(x)}\right)^{\frac{g(x)}{f(x)}}
$$
Claim : there exists a $C$ such that $\frac{g(x)}{f(x)} \to C$ as $x \to \infty$.
Proof : Note that $$
\lim_{x\to \infty} \left(\left(1 + \frac 1{f(x)}\right)^{f(x)}\right)^{\frac{g(x)}{f(x)}} = c
$$
by continuity of the logarithm,
$$
\lim_{x \to \infty}\frac{g(x)}{f(x)}\ln\left(1+\frac{1}{f(x)}\right)^{f(x)} = \ln c
$$
since $\lim_{x \to \infty} \ln\left(1+\frac{1}{f(x)}\right)^{f(x)} = 1$, applying the quotient rule tells us that $\frac{g(x)}{f(x)} \to \ln c$ as $x \to \infty$.
Now, we note that:
$$
\left(1 - \frac 1{f(x)}\right)^{g(x)} = \left(\left(1 - \frac 1{f(x)}\right)^{f(x)}\right)^{\frac{g(x)}{f(x)}}
$$
It is clear that the LHS is always positive, therefore we take $\ln$ on both sides :
$$
\ln\left(1 - \frac 1{f(x)}\right)^{g(x)} = \frac{g(x)}{f(x)}\ln\left(\left(1 - \frac 1{f(x)}\right)^{f(x)}\right)
$$
We know that $\lim_{x \to \infty}\ln\left(\left(1 - \frac 1{f(x)}\right)^{f(x)}\right) = -1$. We also know that $\frac{g(x)}{f(x)} \to \ln c$ as $x\to \infty$. It follows that $$
\frac{g(x)}{f(x)}\ln\left(\left(1 - \frac 1{f(x)}\right)^{f(x)}\right) \to -\ln c
$$
as $x \to \infty$, and therefore by the continuity of the exponential,
$$
\lim_{x\to \infty} \left(1 - \frac 1{f(x)}\right)^{g(x)} = e^{-\ln c} = \frac 1c
$$
as desired.

Note that a stronger converse-like statement is available : if $\frac{g(x)}{f(x)} \to C$ (for any $C \in \mathbb R$) then both the limits $$
\left(1+\frac{1}{f(x)}\right)^{g(x)} \to e^C \\
\left(1-\frac{1}{f(x)}\right)^{g(x)} \to e^{-C}
$$
hold. These follow from the key identity.
I would expect limsup/liminf relations , if available for $\frac{g(x)}{f(x)}$ to translate to $\left(1+\frac 1{f(x)}\right)^{g(x)}$ as well.
A: As David mentions in the comments, we need an additional condition, namely
$$\lim_{x\to \infty} \frac {g(x)}{f(x)} \text{ exists}$$
So,
$$\lim_{x\to \infty}\left(1+\frac1{f(x)}\right)^{g(x)}=\left(\left(1+\frac1{f(x)}\right)^{f(x)}\right)^{\frac{g(x)}{f(x)}}=e^{\frac{g(x)}{f(x)}}$$
and
$$\lim_{x\to \infty}\left(1-\frac1{f(x)}\right)^{g(x)}=\left(\left(1-\frac1{f(x)}\right)^{f(x)}\right)^{\frac{g(x)}{f(x)}}={\left(\frac 1e\right)}^{\frac{g(x)}{f(x)}}=\frac 1{\lim_{x\to \infty}\left(1+\frac1{f(x)}\right)^{g(x)}}$$
This completes the answer.
