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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.

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  • $\begingroup$ What's the question? $\endgroup$
    – zahbaz
    Oct 27, 2016 at 6:34
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    $\begingroup$ I 'm looking for a proof @zahbaz $\endgroup$ Oct 27, 2016 at 6:39

5 Answers 5

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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.

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    $\begingroup$ This works when the limits both exist, since $\exp$ and $\log$ are both continuous. (Phrase $\lim r^s$ as $\lim \exp(s \log r)$, and use that the limit of a product is the product of the limits.) $\endgroup$ Apr 3, 2018 at 10:51
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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, $$ \lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right) $$

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    $\begingroup$ How did you get $\lim_{x\rightarrow a} \log(f)/(f-1) = 1$ in the 4th step? I used l'Hopital's to verify it, but often this formula is taught to students before they see derivatives, so I'm wondering if it can be proved without resort to calculus?! $\endgroup$
    – Fixee
    May 11, 2019 at 2:20
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    $\begingroup$ Without calculus, we are limited in the ways to define $\log(x)$. One way is to define $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$. Then Bernoulli's Inequality says not only that $e^x\ge1+x$, but also that $e^{-x}\ge1-x$ which implies that $e^x\le\frac1{1-x}$. Substituting $x\mapsto\log(x)$ and rearranging gives, for $x\gt0$, $$\frac{x-1}x\le\log(x)\le x-1$$ which says that $$\min\left(\tfrac1x,1\right)\le\frac{\log(x)}{x-1}\le\max\left(\tfrac1x,1\right)$$ $\endgroup$
    – robjohn
    May 11, 2019 at 23:32
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$$\begin{align} \lim \limits_{x \to a} f^g = e^{\lim \limits_{x \to a} g\cdot \log[1+ (f-1)]} &= e^{\lim_{x \to a} g\cdot [(\frac{(f-1)}{1}) + (\frac{(f-1)^2}{2}) + (\frac{(f-1)^3}{3}) + ...]} \\ &= e^{\lim \limits_{x \to a} g\cdot (f-1)[1 + (\frac{(f-1)}{1}) + (\frac{(f-1)^2}{2}) + ...]}. \end{align}$$

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:

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

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    $\begingroup$ I don't know how to write all that math here, though I tried hard to copy and paste these expressions with necessary edits from the tutorial page. This is my first answer, any help regarding editing will be appreciated. $\endgroup$ Apr 3, 2018 at 7:14
  • $\begingroup$ Thank you @Daniel Fischer. $\endgroup$ Apr 3, 2018 at 18:03
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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 $\infty$.

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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: $$\lim_{x\rightarrow a}f^g=\lim_{x\rightarrow a}e^{\frac{\ln(1+f-1)}{f-1}\cdot g(f-1)}=e^{\lim\limits_{x\rightarrow a}\left(\frac{\ln(1+f-1)}{f-1}\cdot g(f-1)\right)}=e^{\lim\limits_{x\rightarrow a}g(f-1)}.$$

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  • $\begingroup$ I belive there is no need in the assumtion that $f \neq 1$ in a neirbourhood of $a$. See Daniel Fischer's proof. $\endgroup$ Aug 9, 2021 at 19:56
  • $\begingroup$ @Enzo Giannotta Maybe. It not says that my reasoning is wrong. By the way, why does in the solution there exist $\lim\limits_{x\rightarrow a}g(f-1)$? It's not given! $\endgroup$ Aug 9, 2021 at 21:06

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