So I understand this is very similar to another question How to prove $\lim_{n \to \infty} (1+1/n)^n = e$?.
But I want to show that for all $x\in\mathbb{R}$, $\lim_{n\to\infty}(1+x/n)^n$ exists. Can I show that this sequence is dominated by $\sum_{k=0}^n \frac{x}{k!}$?
For $x > 0$, it is staightforward to find an upper bound using the binomial formula and then to show that the sequence is monotonically increasing. Then you can show that
$$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n \leqslant e^x,$$
but some additional work is needed to show equality.
I will give you an alternative that handles the case $x < 0$ and can be modified to prove the case with $x > 0$ as well. It also demonstrates uniform convergence.
To prove
$$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n = e^x\,\,\,\, (x <0),$$
consider the following argument to prove the equivalent
$$\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^n = e^{-x}\,\,\,\, (x > 0).$$
Using the inequality $\ln(1+y) \leqslant y$, we have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} < \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} < \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x < \left(1 - \frac{x}{n}\right)^{-n}.$$
Hence, using the Bernoulli inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n,$
$$0 \leqslant e^{-x} - \left(1- \frac{x}{n}\right)^n = e^{-x}\left[1 - e^x\left(1- \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{-x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{-x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{-x}\frac{x^2}{n}.$$
Therefore,
$$0 \leqslant\sup_{x \in [0,\infty)} \left|e^{-x} - \left(1- \frac{x}{n}\right)^n\right| \leqslant \sup_{x \in [0,\infty)}e^{-x}\frac{x^2}{n}= \frac{4e^{-2}}{n}\xrightarrow[n \rightarrow \infty]\quad 0.$$
\begin{align} \lim \limits_{n \to \infty}(1+\frac{x}{n})^n&=\lim \limits_{n \to \infty}((1+\frac{1}{\frac{n} {x}})^{\frac{n} {x}})^x \\ &=\lim \limits_{y \to \infty}((1+\frac{1}{y})^{y})^x \\ &=e^x \end{align} By taking log, it can be proved that $\lim \limits_{y \to \infty}(1+\frac{1}{y})^{y}=e$.
It is also based on fact that $x^r (r \in \mathbb{R})$ is a continuous function of $x$.
