# Does $1^{\infty}=e$ or $1^{\infty}=1$?

In fact the real question is: Does $\lim\limits_{n\to\infty}1^{n}=e$?.

I know that

$$\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e,$$

So, can we say that $1^\infty=e$?

And, by logic, This product $$\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},$$ gives $1$ not other value.

• Perhaps I'm overlooking something but I'm pretty sure $1^n=1$ for all $n$. So no, it does not equal $e$. Also I'm not sure how you are defining $1^\infty$ in a meaningful way. – Nap D. Lover Dec 14 '15 at 19:40
• Writing a limit as $1^{\infty}$ is most often used as a shorthand to indicate that it is an indeterminant form and so you must apply some form of L'Hospital's rule. It is not meant to be literally interpreted as raising $1$ to the $\infty$ power. – Cameron Williams Dec 14 '15 at 19:40
• I think your reasoning error is that you forget that the $n$ is both inside the parenthesis and outside. What you seem to think is that you can first calculate the limit inside and then outside. Which is completely wrong. – PAM Dec 14 '15 at 19:41
• That's why I was confused, for there was no situation of $1^{h(x)}$ where $h(x)\to \infty$ (as $x$ approaches something, whether a point or infinity) in the posed question. – Nap D. Lover Dec 14 '15 at 19:43
• When it comes to limits, you could only substitute the destination value only if the function is defined and continuous at the limit destination point. You can't apply infinity here using substitution and get correct result since the function is NOT defined at infinity, that is f(infinity) is undefined. – NoChance Dec 14 '15 at 19:52

You cannot apply the limit for just one term of an expression at once and then manipulate the rest of it.

Doing so will lead to all kinds of spurious proofs and incorrect results.

The sequence $\{1^n\}_{n \in \Bbb{N}}$ is constant at $1$. Therefore, its limit is 1.

Maybe your confusion comes from the fact that we cannot "add limits". I.e., $$e=\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n \ne \lim_{k\to \infty}\left(\lim_{n\to \infty}\left(1+\frac{1}{n}\right)\right)^k=1.$$

• $2^n \cdot \sin(\pi \cdot n)$ is also constant for $n \in \mathbb{N}$ but $\lim_{n \rightarrow \infty}{2^n \cdot \sin(\pi \cdot n)} \ne 0$ – Dan Brumleve Dec 14 '15 at 19:43
• @DanBrumleve I'm confused now. How can a constant sequence not converge to the constant ? – Nitrogen Dec 14 '15 at 19:47
• @Nitrogen The point is that when we write "$\lim_{n\rightarrow\infty}$," unless otherwise specified, non-integral values of $n$ are also taken into account. (There's a context problem - whether we're talking about the limit of a sequence, or the limit in a space. Arguably, using the variable $n$ suggests that we're working in the type of integers, but technically we should explicitly say that somewhere.) – Noah Schweber Dec 14 '15 at 19:47
• @NoahSchweber: Well, that's a matter of convention. I would say that conventionally, if you're calling the variable "$n$", then by default you are talking about the limit of a sequence. Of course, there is no harm in being more explicit about it. – Eric Wofsey Dec 14 '15 at 19:50
• @Nitrogen No, there's no type theory here - I was using the word "type" informally. Think of the difference between "$\lim_{x\rightarrow\infty}\sin(x)=0$," which is clearly false, versus "the limit of the sequence $\sin(n\pi)$ as $n\rightarrow\infty$ is $0$," which is clearly true. The two expressions differ in the type (informal usage) of object being considered: in the former, $x$ ranges over reals, and in the latter, $n$ ranges over naturals. As Eric quite rightly says, in normal usage this is distinguished (as I have) by choice of variable - $x$ vs. $n$ - (cont'd) – Noah Schweber Dec 14 '15 at 19:55

The situation is actually worse than you think: the following statements are all true: $$\lim_{n \to \infty} 1^n = 1$$ $$\lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^n = e$$ $$\lim_{n \to \infty} \left(1 + \frac{\ln (17)}{n}\right)^n = 17$$ $$\lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^n = e^r$$ Are we to conclude from this that $1^\infty$ equals $1$, $e$, $17$, and $e^r$ for every $r$? Of course not. The point here is that $1^\infty$ does not make sense.

(A more intellectually honest reason why $1^\infty$ does not make sense is that $\infty$ is not a real number, so you cannot expect to manipulate it via exponents as you would with real numbers.)

More generally, as PAM points out in the comments, it is not always true that $$\lim_{n \to \infty} f(n)^{g(n)} = (\lim_{n \to \infty} f(n))^{\lim_{n \to \infty} g(n)}.$$ That is, you can't always evaluate the limits separately and just plug in. (Sometimes you can, but not always.)

So, can we say that $1^\infty=e$?

No. There is no element $\infty$ in $\mathbb{R}$ (or $\mathbb{N}$, for that matter), so usually when you encounter an expression like ${1 \over 2}^{+\infty}=0$ it means "for whatever sequence $a_n : \lim_{n \to \infty}a_n=+\infty$, $\lim_{n\to \infty}{1 \over 2}^{a_n}$ is defined and equals $0$". It's just a short notation about limits, not actual arithmetic.

Now, for $a_n : \lim_{n \to \infty}a_n = \infty, b_n : \lim_{n \to \infty}b_n=1$, we can try to determine what $\lim_{n \to \infty}b_n^{a_n}$ equals to. But, just as you showed, there are two pairs $(a_n, b_n)$ (namely, $(1,n)$ and $(1+{1 \over n},n)$) for which you get different results. That means no general statement can be done in form $1^\infty=X$, this is called an indeterminate form and you have to analyze $a_n$ and $b_n$ to get an answer in each particular case.

Neither. $1^{\infty}$ is called an indeterminate form, clearly indicating that it has no specific value. It is a conventional expression, as $\infty$ is not a number and you cannot perform arithmetic with it.

Computing a limit is something different, it uses the notion of continuity: if you can compute the values of a function for arguments as large as you want, this value can get closer and closer to a given number.

The case of $1^n$ is easy: as $f(n)=1$ for all $n$, the limit is simply $1$.

The case of $\left(1+\dfrac1n\right)^n$ is different. If you compute successive values, you will observe a convergence:

$$\begin{matrix}n&1^n&\left(1+\frac1n\right)^n\\ 1&1&2\\ 10&1&2.59\cdots\\ 100&1&2.705\cdots\\ 1000&1&2.71692\cdots\\ 10000&1&2.718146\cdots\\ \end{matrix}\\\cdots$$

One can rigorously prove that there is a limit, which is an irrational number that can be quickly evaluated as

$$e=\sum_{k=0}^\infty \frac1{k!}.$$ This sum is another sequence that has a limit $\approx2.7182818284590452353602874713527\cdots$

On the opposite, $(1+\frac1{\sqrt n})^n$ grows approximately like $e^{\sqrt n}$ and has no limit.

Congratulations, you have discovered a false belief.

False belief. Insertion of limits "rule" $$\lim_{x \rightarrow x_0}f(g(x),x) = \lim_{x \rightarrow x_0}f\left(\lim_{y \rightarrow x_0}g(y),x\right)$$

Counterexample.

$$f(a,b) = \left(1+a\right)^b \qquad g(a)=\frac{1}{a}$$

Then $$\lim_{x \rightarrow \infty}f(g(x),x) = e, \qquad \lim_{x \rightarrow \infty}f\left(\lim_{y \rightarrow \infty}g(y),x\right) = 1$$

You are mixing up two things. First lets get to the simpler matter. When you see an expression of type $a^{b}$ (or $ab$ or $a/b$ or $a \pm b$), before you figure out the meaning of that expression it is essential to figure out what are $a$ and $b$ and then ask if the corresponding operation for those things $a, b$ is defined or not.

Normally in such expression $a, b$ are either real or complex numbers and then these operations are defined in a particular manner with certain restrictions. Thus for example $a/b$ is defined for all complex numbers $a, b$ except $b = 0$. Similarly there are certain restrictions for the definition of expression $a^{b}$. In your current case $1^{\infty}$ is something meaningless simply because $\infty$ is neither a real nor a complex number.

Second mistake is about the notion of limit. A limit of a sequence $\{s_{n}\}$ is defined in a very specific manner and there are rules to calculate limits of sequences. Based on these definitions and rules it follows that $$\lim_{n \to \infty}1^{n} = 1\text{ and }\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ Out of these the first limit is easy to explain while the second limit is not so easy to explain. For the first limit we know that $1^{n} = 1$ for all natural numbers $n$ and hence the sequence is constant with a value $1$ and therefore its limit is also $1$.

For the second limit it can be shown with some difficulty that it exists and lies between $2$ and $3$. Most textbooks define this limit to be $e$ (and some other textbooks define $e$ in somewhat different manner and show that it is also the value of this limit).

Moreover I can understand the line of thought behind expressions like $1^{\infty}$. Beginners almost always think that a limit is obtained just by "plugging" (e.g $\lim_{x \to 1}(x + 1) = 1 + 1 = 2$). However this is very very wrong. A limit is evaluated by using definition of limit and certain rules to evaluate limits and not by plugging. Moreover when we deal with limit of sequences this plugging simply does not work because the limit is always $n \to \infty$ and the symbol $\infty$ is not a number which can be plugged in place of $n$. So it is very very wrong to go from $1^{n}$ to $1^{\infty}$.