Regarding $e$ in $\lim\limits_{x \to a}{[\phi(x)]^{\psi(x)}} = e^{\lim\limits_{x \to a}{[\phi(x) - 1]\psi(x)}}$

I'm currently studying limits with $x$ in the exponent. The following formula simplifies the work to solve limits.

If $\lim\limits_{x \to a}{\phi(x)} = 1$ and $\lim\limits_{x \to a}{\psi(x)} = \infty$, then

$\lim\limits_{x \to a}{[\phi(x)]^{\psi(x)}} =$ $\lim\limits_{x \to a}{\{[1+\alpha(x)]^{\frac{1}{\alpha(x)}}\}^{\alpha(x)\psi(x)}} =$ $e^{\lim\limits_{x \to a}{[\phi(x) - 1]\psi(x)}}$

For the most part, I understand how the formula is derived. However, there's one part I don't understand.

Why does $\lim\limits_{x \to a}{\{[1+\alpha(x)]^{\frac{1}{\alpha(x)}}\}} = e$?

METHODOLOGY $1$: Pre-Calculus Approach

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

Letting $x$ be replaced with $1+\alpha(x)$ in $(1)$ reveals

$$\frac{\alpha(x)}{1+\alpha(x)}\le \log(1+\alpha(x))\le \alpha(x)\tag 2$$

Then, noting that $(1+\alpha(x))^{1/\alpha(x)}=e^{\frac{1}{\alpha(x)}\log(1+\alpha(x))}$ and applying $(2)$, we find that

$$e^{\frac{1}{1+\alpha(x)}}\le (1+\alpha(x))^{1/\alpha(x)}\le e \tag3$$

whence applying the squeeze theorem to $(3)$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to a}(1+\alpha(x))^{1/\alpha(x)}=e}$$

METHODOLOGY $2$: Asymptotic Analysis

We have for $\alpha(x) \to 0$ as $x\to a$

$$\begin{align} (1+\alpha(x))^{1/\alpha(x)}&=e^{\frac{1}{\alpha(x)}\color{blue}{\log(1+\alpha(x))}} \\\\ &=e^{\frac{1}{\alpha(x)}\color{blue}{\left(\alpha(x)+O\left(\alpha^2(x)\right)\right))}}\\\\ &=e^{1+O(\alpha(x))}\\\\ &\to e\,\,\text{as}\,\,x\to a \end{align}$$

In the development herein, we used the asymptotic expansion $\displaystyle \log(1+t)=t+O(t^2)$.