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I know infinity is not a number so is this question undefined or something Can anyone explain it to me please
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.
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, $$ \lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right) $$
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.
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, $$ \lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right) $$
It isn’t: $\lim_{n\to\infty}1^n=1$, exactly as you suggest. However, if $f$ and $g$ are functions such that $\lim_{n\to\infty}f(n)=1$ and $\lim_{n\to\infty}g(n)=\infty$, it is not necessarily true that
$$\lim_{n\to\infty}f(n)^{g(n)}=1\;.\tag{1}$$
For example, $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\approx2.718281828459045\;.$$
More generally,
$$\lim_{n\to\infty}\left(1+\frac1n\right)^{an}=e^a\;,$$
and as $a$ ranges over all real numbers, $e^a$ ranges over all positive real numbers. Finally,
$$\lim_{n\to\infty}\left(1+\frac1n\right)^{n^2}=\infty\;,$$
and
$$\lim_{n\to\infty}\left(1+\frac1n\right)^{\sqrt n}=0\;,$$
so a limit of the form $(1)$ always has to be evaluated on its own merits; the limits of $f$ and $g$ don’t by themselves determine its value.
The limit of $1^{\infty}$ exist:$$\lim_{n\to\infty}1^n$$ is not indeterminate. However$$\lim_{a\to 1^+,n\to\infty}a^n$$ is indeterminate..