Finding exponential function limit definition from definition of e I am trying to prove, from the following limit definition of $e$
$$e=\lim_{n\to+\infty} \left(1+\frac{1}{n}\right)^n$$
the following definition for the exponential function:
$$e^x =\lim_{n\to+\infty} \left(1+\frac{x}{n}\right)^n$$
I tried to follow the answer of this post, but there is something I don't understand: Prove $e^x$ limit definition from limit definition of $e$.
Here is a screenshot:

To substitute $n$ for $u$ in the limit "index", we need to make sure that $u$ goes to $+\infty$ as $n$ goes to $+\infty$.
Therefore, we should have:
$$\lim_{n\to+\infty} u = \lim_{n\to+\infty} nx \stackrel{?}{=}+\infty$$
Which is, in some way, $(+∞)*x$.
That is fine if $x$ is positive, but what if $x$ were negative? Wouldn't that limit then be equal to $(-∞)$, which would invalidate that "re-indexing"?
Is the proof missing something, or am I wrong?
Thanks in advance!
(The reason I am writing this here and not commenting is that the site settings forbid me to, because of my limited activity.)
 A: I will assume exponentiation is continuous. Let $f(x)=\lim_{n\to\infty}(1+\frac xn)^n$. For $x>0$, I will show $e^x=f(x)$. [The case when $x<0$ is similar.]
Lemma $f(x)$ is continuous.
Proof I will prove it is continuous at $x=x_0$, for arbitrary $x_0\in\mathbb R$. For all $\epsilon>0$, let $\delta=\epsilon/f(|2x_0|)$. Let $x\in\mathbb R$ be such that $|x-x_0|<\delta$. Then,
\begin{align}
|f(x)-f(x_0)|&=\lim_{n\to\infty}\left|\big(1+\frac xn\big)^n-\big(1+\frac {x_0}n\big)^n\right|\\
&=\lim_{n\to\infty}\left|\sum_{i=1}^n{n\choose i}\frac{x^i-x_0^i}{n^i}\right|\\
&<\delta\lim_{n\to\infty}\sum_{i=1}^n{n\choose i}\frac{i|2x_0|^{i-1}}{n^i}.
\end{align}
Here, the sum in the limit is
$$
\sum_{i=1}^n{n\choose i}\frac{i|2x_0|^{i-1}}{n^i}=\sum_{i=1}^n{n-1\choose i-1}\frac{|2x_0|^{i-1}}{n^{i-1}}\to f(|2x_0|) \ (n\to\infty).
$$
Thus, we obtain $|f(x)-f(x_0)|<f(|2x_0|)\delta=\epsilon$. QED
Thus, since both $e^x$ and $f(x)$ are continuous, it suffices to check they are equal for rational numbers $x=p/q$ with $p,q$ positive integers.
It is well-known that taking a subsequence does not change the limit of a convergent sequence. Thus,
\begin{align}
e^{p/q}&=\lim_{n\to\infty}\big(1+\frac1n\big)^{pn/q}\\
&=\lim_{n\to\infty}\big(1+\frac1{qn}\big)^{pn},
\end{align}
and
\begin{align}
f(p/q)&=\lim_{n\to\infty}\big(1+\frac{p}{qn}\big)^n\\
&=\lim_{n\to\infty}\big(1+\frac1{qn}\big)^{pn},
\end{align}
equal each other.
A: If you don't have any restrictions on how to prove, then you can use logarithm method-
Let $f(n) = (1 + \frac{1}{n})^{n}$. Then,
$\lim\limits_{n \to \infty} f(n) = \lim\limits_{n \to \infty} (1 + \frac{1}{n})^{n} = \Large e^{\big[\large \lim\limits_{n \to \infty} \large\ln \big(1 + \frac{1}{n}\big)^{n}\big]}$
So, instead of finding $\lim\limits_{n \to \infty} f(n)$, try to find $\lim\limits_{n \to \infty} \ln(f(n))$.
So, $\lim\limits_{n \to \infty} \ln(f(n)) = \lim\limits_{n \to \infty} \ln \big(1 + \frac{1}{n}\big)^{n}$
$= \lim\limits_{n \to \infty} n\ln \big(1 + \frac{1}{n}\big) = \lim\limits_{n \to \infty} \frac{\ln \big(1 + \frac{1}{n}\big)}{1/n}$
Now this is $\frac{0}{0}$ form, so using L'Hôpital's rule,
$\lim\limits_{n \to \infty} \ln(f(n)) = \lim\limits_{n \to \infty} \frac{1}{1+1/n}*\big(\frac{-1}{n^{2}}\big)*\left(\Large\frac{1}{\frac{\Large -1}{\Large n^{2}}}\right)$
$= \lim\limits_{n \to \infty} \frac{1}{1+1/n} = 1$
So, $\lim\limits_{n \to \infty} f(n) = e^{1} = e$
Same can be used for proving $\lim\limits_{n \to \infty} (1 + \frac{x}{n})^{n} = e^{x}$
