# 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.

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

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.

• 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.) Apr 3, 2018 at 10:51

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, $$\lim_{x\to a}f^g=\exp\left(\lim\limits_{x\to a}\,(f-1)\,g\right)$$

• 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?! May 11, 2019 at 2:20
• 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 $$\min\left(\tfrac1x,1\right)\le\frac{\log(x)}{x-1}\le\max\left(\tfrac1x,1\right)$$
– robjohn
May 11, 2019 at 23:32

\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:

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

• 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. Apr 3, 2018 at 7:14
• Thank you @Daniel Fischer. Apr 3, 2018 at 18:03

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 $$\infty$$.

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

• I belive there is no need in the assumtion that $f \neq 1$ in a neirbourhood of $a$. See Daniel Fischer's proof. Aug 9, 2021 at 19:56
• @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! Aug 9, 2021 at 21:06