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.

Question

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 $x$.

Answer 1

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 $e$.

Answer 2

Hint:

$$ \lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n \lim_{n\to\infty} \left( 1-\frac{x}{n} \right)^n = \lim_{n\to\infty} \left( 1-\frac{x^2}{n^2} \right)^n = \lim_{n\to\infty} \left( 1-\frac{x^2}{n^2} \right)^{n^2/n }=1.$$

Answer 3

If $x<0$ then $-x>0$. Now apply it for $-x$ and make the substitution