For $x\in\mathbb{R}$ and $n\in\mathbb{N}$, define the real sequence: $$a_n:=(1+x/n)^n$$ Now prove that there exists an $\alpha\in\mathbb{R}$ (just prove that $\alpha$ exists, no need to find the value) such that: $$\forall\epsilon>0,\exists N\in\mathbb{N}:\forall n\ge N,|a_n-\alpha|<\epsilon$$
This should be done strictly with basic algebra; no logarithms, no Taylor series, no derivatives, and $e$ shouldn't appear anywhere in the proof. I suspect that the Bernoulli inequality will probably be useful. Considering the limitations, is there a concise way to show this?
MONOTONICITY
In THIS ANSWER, I showed using Bernoulli's Inequality that $\left(1+\frac xn\right)^n$ is monotonically increasing for $x>-n$.
BOUNDED ABOVE
Using the binomial theorem, we have for $x>-n$
$$\begin{align} 0&\le\left(1+\frac xn\right)^n\\\\ &=\left|\sum_{k=0}^n\binom{n}{k}\left(\frac xn\right)^k\right|\\\\ &=\left|\sum_{k=0}^n \prod_{\ell=1}^{k-1} \left(1-\frac{\ell}{n}\right)\frac{x^k}{k!}\right|\\\\ &\le \sum_{k=0}^n \frac{|x|^k}{k!}\\\\ &\le \sum_{k=0}^\infty \frac{|x|^k}{k!} \end{align}$$
where the ratio test guarantees that the series $\sum_{k=0}^\infty \frac{|x|^k}{k!}$ converges for every $x$.
CONCLUSION:
Inasmuch as $\left(1+\frac xn\right)^n$ is bounded above and monotonically increasing for each fixed $x$, then it converges for each fixed $x$.
And we are done!