1 to the power of infinity formula There is a general formula for indeterminate form $1 ^ {\infty}$ which I'm looking for a proof which is also used here. (picture)

Given $$\lim_{x\to a} f(x) = 1$$ and $$\lim_{x\to a} g(x) = \infty$$, what is
$$\lim_{x\to a} f^{g} = e^{\lim_{x\to a}{(f-1)g}}\quad ? $$

I would appreciate it if somone could give me a proof of this formula.
 A: $$\begin{align}
\lim \limits_{x \to a} f^g = e^{\lim \limits_{x \to a} g\cdot \log[1+ (f-1)]}
&= e^{\lim_{x \to a} g\cdot [(\frac{(f-1)}{1}) + (\frac{(f-1)^2}{2}) + (\frac{(f-1)^3}{3}) + ...]} \\
&= e^{\lim \limits_{x \to a} g\cdot (f-1)[1 + (\frac{(f-1)}{1}) + (\frac{(f-1)^2}{2}) + ...]}.
\end{align}$$
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:
$$\lim_{x \to a} f^g = e^{\lim \limits_{x \to a} g\cdot (f-1)}.$$
A: 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.
A: Remember that $f$ and $g$ are functions of $x$, so to be more precise, we should write $f(x)$ and $g(x)$ instead of $f$ and $g$. This applies to the answer below and to the other answers which have also adopted the shorthand used in the question of $f$ for $f(x)$ and $g$ for $g(x)$.

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,
$$
\lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right)
$$
A: Some limits are indeterminate because, depending on the context, they can evaluate to different ends.  For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers.
\begin{align}\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n &= 1 \\
\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n &= e\\
\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n &= \infty \\
\end{align}
Limits are entirely concerned with the journey of how the approach is taken. I could envoke Robert Frost here (two paths diverged in a wood...), suffice to say take any number (even by a $\varepsilon$) larger than 1, raise this to an arbitrarily large nymber, and the journey will head to $\infty$.
A: It's wrong in the general.
It's true if there exist $\lim\limits_{x\rightarrow a}g(f-1)$ and 
there is $\delta>0$ for which $f\neq1$ for any $0<|x-a|<\delta$ we have $f(x)\neq1$.
Indeed, since $h(x)=e^x$ is a continues function we obtain:
$$\lim_{x\rightarrow a}f^g=\lim_{x\rightarrow a}e^{\frac{\ln(1+f-1)}{f-1}\cdot g(f-1)}=e^{\lim\limits_{x\rightarrow a}\left(\frac{\ln(1+f-1)}{f-1}\cdot g(f-1)\right)}=e^{\lim\limits_{x\rightarrow a}g(f-1)}.$$
