Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$? To prove $\lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$ exists, we prove that the sequence
$$f_n=\left(1+\frac{x}{n}\right)^n$$
is bounded and monotonically increasing toward that bound.
Proof Attempt:

We begin by showing $f_n=\left(1+\frac{x}{n}\right)^n$ is monotonically increasing by looking at the ratio of consecutive terms:
\begin{align*}
\frac{f_{n+1}}{f_n}
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n}} \tag{Definition of $f_n$} \\
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)^{n}\left(1+\frac{x}{n}\right)} \tag{Multiplication by $\frac{\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)}$} \\
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n+1}}\left(1+\frac{x}{n}\right) \tag{Simplify $a^n\cdot a = a^{n+1}$} \\
&=\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a^{n+1}}{b^{n+1}}=\left(\frac{a}{b}\right)^{n+1}$} \\
&=\left(\frac{\frac{n+1+x}{n+1}}{\frac{n+x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Common denominators} \\
&=\left(\frac{n+1+x}{n+1}\cdot \frac{n}{n+x}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$} \\
&=\left(\frac{n^2+n+nx}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Distribute $(n+1+x)n$} \\
&=\left(\frac{n^2+n+nx+x-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Add and subtract $x$} \\
&=\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Factor $n^2+n+nx+x$} \\
&=\left(1+\frac{-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$} \\
&\ge\left(1+\frac{-x}{(n+x)}\right)\left(1+\frac{x}{n}\right) \tag{Bernoulli: $(1+x)^n \ge 1+nx$} \\
&=\left(\frac{n}{n+x}\right)\left(\frac{n+x}{n}\right) \tag{Common denominators} \\
&=1 \tag{Simplify $\frac{a}{b} \cdot \frac{b}{a}=1$}
\end{align*}
Since $\frac{f_{n+1}}{f_n}>1$, then $f_{n+1}>f_n$, which shows the sequence $f_n$ is monotonically increasing for all $n \in \mathbb{N}$.
Next, we show $f_n=\left(1+\frac{x}{n}\right)^n$ is bounded above. Note that
\begin{align*}
f_n
&=\left(1+\frac{x}{n}\right)^n \tag{Definition of $f_n$} \\
&=\sum_{k=0}^n \binom{n}{k} (1)^{n-k} \left(\frac{x}{n}\right)^{k} \tag{Binomial Theorem} \\
&=1+\frac{n}{1!}\left(\frac{x}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{x}{n}\right)^3+\cdots+\frac{n!}{n!}\left(\frac{x}{n}\right)^n  \\
&=1+\frac{\frac{n}{n}}{1!}x+\frac{\frac{n(n-1)}{n^2}}{2!}x^2+\frac{\frac{n(n-1)(n-2)}{n^3}}{3!}x^3+\cdots+\frac{\frac{n!}{n^n}}{n!}x^n \tag{Simplify}\\
&=1+\frac{1}{1!}x+\frac{\left(1-\frac{1}{n}\right)}{2!}x^2+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}{3!}x^3+\cdots+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)}{n!}x^n \\
& \le 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots+\frac{1}{n!}x^n \tag{$1-\frac{k}{n}<1$} \\
& = \sum_{k=0}^n \frac{1}{k!} x^k \tag{Sigma notation}\\
& \to %\underset{n \to \infty}{\to} 
\sum_{k=0}^\infty \frac{1}{k!}x^k \tag{as $n \to \infty$} \\
& = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{k!}x^k \tag{$\exists K$, $k>K$ implies $k! \ge (2x)^k$}\\
& \le \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{(2x)^k} x^k \tag{$k! \ge (2x)^k$ implies $\frac{1}{k!} \le \frac{1}{(2x)^k}$}\\
& = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{2^k} \tag{$\frac{1}{(2x)^k}x^k=\frac{1}{2^k x^k}x^k = \frac{1}{2^k}$}\\
&= \sum_{k=0}^K \frac{1}{k!} x^k + \frac{1}{2^K} \tag{Geometric series evaluation}
\end{align*}
which is finite. Thus, the sequence $f_n$ is bounded. Since it is both monotonically increasing and bounded, it is convergent by the Monotone convergence theorem.

Is my proof correct? I am suspicious of the step which says "$\rightarrow \sum_{k=0}^n \frac{1}{k!}x^k$", and would like to avoid taking another limit in the middle of the boundedness proof.
I also compared my proof to the following references and saw something worrisome:

*

*Reference 1  <- Assumes $x\ge 0$ (Why?)


*Reference 2  <- Assumes $x \ge -1$  (Why?)


*Reference 3  <- Considers $x=0$, $x>0$, and $x<0$ separately (Why?)
All of the above proofs either assumed $x>0$ or considered cases where $x>0$ and $x<0$ separately, but I do not know why. In fact, the third reference considers $\left(1-\frac{x}{n}\right)^{-n}$ for $x>0$ (I think this is a typo and should read $x<0$), but I am not sure why the negative exponent is needed (we are talking about a negative value of $x$, not negative $n$.)
I could only find one proof that did not consider different cases on the sign of $x$:

*

*Bonus reference 4 <- Uses absolute values, but I am not sure why these are necessary either.

I would like to verify my proof and ask 3 questions:

*

*Why is it necessary to consider cases $x>0$ and $x<0$ separately? Did any step in my proof implicitly assume that $x>0$? If so, which one?


*Is there any way to avoid taking a limit in the middle of the boundedness proof?


*Substituting $n=1$ in my boundedness proof shows $1+x \le \sum_{k=0}^n \frac{1}{k!}x^k$. Does this imply $1+x \le \lim\limits_{n\to \infty} \left(1+\frac{x}{n}\right)^n$, since $f_n$ is an increasing function of $n$? Can this be seen explicitly, or would that require a separate proof?
Thank you.

*

*You application of Bernoulli's inequality is correct as long as $y_n=-\frac{x}{(n+x)(n+1)}\geq-1$. In reference (2) which you mention in your posting, Mark Viola assumes $x> -1$, in which case $y_n>-1$ for all $n\geq1$ and so,   the sequence $f_n$ (in your notation) is monotone increasing for all $n\in\mathbb{N}$. However, he also points out that for any $x$, one can consider the first $n_0\in\mathbb{N}$ so that $n_0+x>0$. Then $y_n>-1$ for all $n\geq n_0$ and so, the sequence $f_n$ is monotone increasing for all for $n\geq n_0$. This shows that the limit exists for all $x\in\mathbb{R}$.


*An alternative approach shows that it is enough to consider the existence of the  limit $e=\lim_n(1+\tfrac1n)^n$ to show that $\lim_n(1+\tfrac{a}{n})^n$ exists and equals $e^a$ for all $a\in\mathbb{R}$.
Indeed, from the existence of  $e=\lim_n(1+\tfrac1n)^n$, we have that
$\lim_{x\rightarrow\infty}(1+\tfrac{1}{x})^x$ ($x$ ranges along positive reals) exists and has limit $e$. To see this, notice that
$n_x:=\lfloor x\rfloor\leq x\leq \lfloor x\rfloor +1 = n_x+1$ hence
$$\big(1+\tfrac{1}{n_x+1}\big)^{n_x}\leq\big(1+\tfrac1x\big)^{n_x}\leq\big(1+\tfrac1x\big)^x\leq\big(1+\tfrac{1}{n_x}\big)^x\leq\big(1+\tfrac{1}{n_x}\big)^{n_x+1} 
$$
As $n_x\xrightarrow{x\rightarrow\infty}\infty$, one gets that
$$
\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x+1}\big)^{n_x}=\lim_{n_x\rightarrow\infty}\frac{\big(1+\tfrac{1}{n_x+1}\big)^{n_x+1}}{1+\tfrac{1}{n_x+1}}=e
$$
and
$$
\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x}\big)^{n_x+1}=\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x}\big)^{n_x}\big(1+\tfrac{1}{n_x+1}\big)=e
$$
Therefore $\lim_{x\rightarrow\infty}\big(1+\tfrac1x\big)^x=e$.
As a consequence, for $a>0$, we have
$$\lim_{n\rightarrow\infty}\big(1+\tfrac{a}{n}\big)^n=\lim_{n\rightarrow\infty}\left(\big(1+\tfrac{a}{n}\big)^{\tfrac{n}{a}}\right)^a=e^a
$$
Similarly, for $x<0$, let $y=-x$ so that $y>0$.
$$\big(1+\tfrac{1}{x}\big)^x=\big(1-\tfrac1y\big)^{-y}=\left(\frac{1}{1-\frac1y}\right)^y=\left(1+\tfrac{\tfrac1y}{1-\tfrac1y}\right)^y=\left(1+\tfrac{1}{y-1}\right)^{y-1}\big(1+\tfrac{1}{y-1}\big)$$
Thus,
$$\lim_{x\rightarrow-\infty}\big(1+\tfrac1x\big)^x=\lim_{y\rightarrow\infty}\left(1+\tfrac{1}{y-1}\right)^{y-1}\big(1+\tfrac{1}{y-1}\big)=\frac1e$$
As a consequence, for $a>0$
$$\lim_{n\rightarrow\infty}\big(1-\tfrac{a}{n}\big)^n=\lim_{n\rightarrow\infty}\left(\big(1-\tfrac{a}{n}\big)^{-\tfrac{n}{a}}\right)^a=e^{-a}$$

*

*Notice that all these requires a definition of real powers that is a rigorous definition of $a^x$ for $a>0$ and $x\in\mathbb{R}$. This of course is done by starting with rational powers, show monotonicity  and then use axioms such as that of the supremum or its equivalents. Other modern approaches are based on integration. One defines $\ln(x)=\int^x_1\frac{1}{t}\,dt$ for $x>0$ and shows that $\ln$ is a nice continuous, monotone increasing functions (furthermore, it is differentiable by the fundamental theorem of Calculus) taking values in all of $\mathbb{R}$, with the property that $\log(ab)=\log(a)+\log(b)$ for all $a,b>0$. Then, the exponential function is defined as the inverse of $\ln$. Theorems such as the inverse function theorem, would imply differentiability
and the property that $\exp(a+b)=\exp(a)\exp(b)$. The number $e=\exp(1)$.

