1 to the power of infinity formula

I believe this is where the identity is coming from.

\begin{align} \lim_{x\to a}f^g &= \lim_{x\to a}(1+f-1)^g \\ &= \lim_{x\to a}\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^g \\ \\ &= \lim_{x\to a}\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^{g\frac{f-1}{f-1}} \\ \\ &= \lim_{x\to a}\left[\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^{\frac{1}{f-1}}\right]^{g(f-1)} \\ \\ &= \lim_{x\to a}\left[\left(1+\frac{1}{\left(\frac{1}{f-1}\right)}\right)^{\frac{1}{f-1}}\right]^{\lim_{x\to a}g(f-1)} \qquad (*) \\ \\ &= e^{\lim_{x\to a}g(f-1)} \end{align}

Where the first limit is a form of the limit definition of $e$. I put a note (*) next to one step I am uneasy about. I am unsure why we can separately evaluate limits here. Perhaps someone else can comment on this.

If $\lim\limits_{x\to a}f=1$, then $$ \begin{align} \log\left(\lim_{x\to a}f^g\right) &=\lim_{x\to a}\log\left(f^g\right)\\[6pt] &=\lim_{x\to a}\log(f)\,g\\ &=\lim_{x\to a}\frac{\log(f)}{f-1}\lim_{x\to a}\,(f-1)\,g\\[6pt] &=\lim_{x\to a}\,(f-1)\,g \end{align} $$ Therefore, $$ \lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right) $$

\begin{align} \lim \limits_{x \to a} f^g = e^{\lim \limits_{x \to a} g*log[1+ (f-1)]} &= e^{\lim \limits_{x \to a} g*[(\frac{(f-1)}{1}) + (\frac{(f-1)^2}{2}) + (\frac{(f-1)^3}{3}) + ...]} \\ &= e^{\lim \limits_{x \to a} g*(f-1)[1 + (\frac{(f-1)}{1}) + (\frac{(f-1)^2}{2}) + ...]} \end{align}

Now, since $f \to 1$ when $x \to a$, all the subsequent terms in the expansion involving $(f-1)$ will become zero and the expression becomes:

$$\lim \limits_{x \to a} f^g = e^{\lim \limits_{x \to a} g*(f-1)}$$