Proving $\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n = \lim_{n\to\infty} \left( 1+\frac{1}{n} \right)^{nx}$ with a certain method. I saw in another post on the website a simple proof that $$\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n = \lim_{m\to\infty} \left( 1+\frac{1}{m} \right)^{mx}$$
which consists of substituting $n$ by $mx$. I can see how the equality then holds for positive real numbers $x$, yet it isn't obvious to me why it holds for negative $x$.

For $x$ negative, we must substitute $n$ with $-mx$ with $m$ positive. We can use the fact that we know what the limit $(1-\frac{1}{m})^{-m}$ is, and equate this with the usual definition of $e$.
