Is it also true that $\lim_{n\to-\infty} \left(1+\frac{x}{n}\right)^n = e^x$? We know that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x.$$
Can we change the "$n\to\infty$" to "$n\to-\infty$"? That is, is the following also true? $$\lim_{n\to-\infty} \left(1+\frac{x}{n}\right)^n = e^x.$$
 A: Yes, just changing $x$ to $-x$ we get $(1-\frac  x  n)^{n} \to e^{-x}$ as $n \to \infty$. This is same as saying that $(1+\frac   x n)^{-n} \to e^{-x}$ as $n \to -\infty$. Now just take reciprocals.
A: Since $e^x$ and $\ln$ are continuous functions, for $x\neq0$ we obtain: $$\lim_{n\to-\infty} \left(1+\frac{x}{n}\right)^n =\lim_{n\to-\infty} \left(\left(1+\frac{x}{n}\right)^\frac{n}{x}\right)^x =e^{x\ln\lim\limits_{n\rightarrow-\infty}\left(1+\frac{x}{n}\right)^{\frac{n}{x}}}= e^x.$$
A: Clearly we have $$\lim_{n\to-\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty} \left(1-\frac{x}{n}\right)^{-n}\tag{1}$$ Next we will show (later) that $$\lim_{n\to\infty} \left(1-\frac{x^2}{n^2}\right)^{n}=1\tag{2}$$ which is same as $$\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n\left(1-\frac{x}{n}\right)^n=1$$ And this implies $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right) ^n=\lim_{n\to\infty} \left(1-\frac{x}{n}\right)^{-n}\tag{3}$$ And equations $(1)$ and $(3)$ complete the proof.
This is based on the assumption that for every real $x$ the limit $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$$ exists and is non-zero.
Now it remains to prove identity $(2)$. This is mostly easily handled by using Bernoulli's inequality and Squeeze theorem. Since $n\to\infty $ we can safely assume $n^2>x^2$ and therefore by Bernoulli's inequality we have $$\left(1-\frac{x^2}{n^2}\right)^n\geq 1-n\cdot\frac{x^2}{n^2}$$ and thus we have $$1-\frac{x^2}{n}\leq \left(1-\frac{x^2}{n^2}\right)^n\leq 1$$ Now using Squeeze theorem gives us identity $(2)$.
This proof avoids the identity $e^{-x} =1/e^x$ used in another answer here. 
