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If I were given $\lim_{n\to \infty}(1 + \frac{1}{n})^n$, and asked to solve, I would do so as follows:

$$\lim_{n\to \infty}(1 + \frac{1}{n})^n$$ $$=(1 + \frac{1}{\infty})^\infty$$ $$=(1 + 0)^\infty$$ $$=1^\infty$$ $$=1$$

I'm aware that this limit is meant to equal to $e$, and so I ask:

why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$?

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To say it briefly: $1^\infty\neq 1$ – Yurii Savchuk Nov 3 '13 at 18:57
You can't just replace $n$ by $\infty$ and call it solving a limit. – Christoph Nov 3 '13 at 18:58
Why has this got upvotes? All this question shows is you have no idea what a limit is. $\lim{n\to\infty}$ doesn't mean "sub in $\infty$". – bwv869 Nov 3 '13 at 19:14
@OliverBel I find this a good question on the beginner's level. Lots of students find this type of limit calculations confusing/valid. The up votes perhaps reflect the fact that OP is not the only one finding this situation confusing. – Ittay Weiss Nov 3 '13 at 19:41
@OliverBel : maybe becaue the OP took the time to type his efforts in the question. – Stefan Smith Nov 3 '13 at 20:23

10 Answers 10

up vote 19 down vote accepted

The main problem here is that you use $\infty $ as if it were a number, and then you just substitute $\infty $ for $n$ and 'compute'. But, since infinity is not a number, you can't just substitute it and the computation you make is meaningless.

I'm assuming from the way you ask the question that you already know how to derive the correct value and that you are just wondering what is wrong with your approach. So, you don't get $e$ when you compute the limit they way you did precisely because your computation is invalid since you treat infinity as a number.

This may be extra confusing since sometimes substituting $\infty $ does lead to the correct answer, but this should be regarded as a fluke. For instance, the limit $\lim _{n\to \infty }\frac {1}{n}$ is $0$, which is what you would get by substituting $\infty $. But this is just coincidence that a completely faulty line of argument using entirely wrong 'computations' leads to the correct answer. Unfortunately, often when teaching calculus such substitutions and manipulations with $\infty $ are glossed over, or worse even encouraged. It is good practice to never ever substitute $\infty $ and compute with it.

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Hint : $1^\infty=(a^0)^\infty=a^{0\cdot\infty}=a^\text{Indeterminate}=\text{Indeterminate}$.

A more comprehensive list of such undetermined expressions can be found here.

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I would say that $1^\infty $ should be interpreted as $\lim _{n\to \infty }1^n$ (which is $1$) rather then as the indeterminate expression you came up with. – Ittay Weiss Nov 3 '13 at 19:33
The point is that the problem with OP's approach is not that $1^\infty $ is or is not $1$ (and I say, it is $1$). The problem is in the steps leading to the wrong conclusion that the limit OP is interested in is the same as $1^\infty$. The problem stems from substituting $\infty $ for $n$, a practice that should never be done. – Ittay Weiss Nov 3 '13 at 19:55
But it isn't... Pretty much for the same reason for which $0\cdot\infty$ cannot be reduced to the above mentioned case either, but is considered indeterminate. For instance, $1^\infty$ can also be viewed as $\lim_{n\to1}n^\infty$ , which can be either $0$ or $\infty$ , depending on how $n$ approaches the value $1$ . – Lucian Nov 3 '13 at 20:02
things like $\lim _{n\to \infty }n^\infty $ have no meaning at all. You seem to interpret the expression $1^\infty $ as the limit of something that tends to $1$ raised to the power of something that tends to $\infty $, and that of course is not uniquely defined. But a much more natural interpretation for $1^\infty $ is as the limit of $1$ raised to the power of something that tends to $\infty $, and that is simply $1$. Anyway, this is besides the point of OP's question, and is just a matter of how does one interpret $1^\infty $. – Ittay Weiss Nov 3 '13 at 20:05
The point is that the interpretation and/or value of $1^\infty $ is only one fragment of where OP erred. The initial reduction of the limit OP was computing to the expression $1^\infty $ is where the problems start. Substituting $\infty $ and playing with it as if it were a number is a practice that should never be done. – Ittay Weiss Nov 3 '13 at 21:36

Consider $$\lim_{n\rightarrow \infty}(1+{1\over n})^n.$$ This limit has the indeterminate form $1^\infty$. Let $y=(1+{1\over n})^n$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\ln(y)=n\ln(1+{1\over n})$. The $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}n\ln(1+{1\over n})$$ which has the indeterminate form $\infty\cdot 0$. We can rewrite the right-hand side limit as $$\lim_{n\rightarrow \infty}\ln(y)={\lim_{n\rightarrow \infty}={\ln(1+{1\over n})\over {1\over n}}}$$ which has the indeterminate form ${0\over 0}$. Using L'Hospital's Rule we see that $$\lim_{n\rightarrow \infty}\ln(y)=\lim_{n\rightarrow \infty}{{1\over (1+{1\over n})}\cdot {-1\over n^2}\over {-1\over n^2}}.$$ This simplifies to $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}(1+{1\over n})=1.$$ So far we have computed the limit of $\ln(y)$, what we really want is the limit of $y$. We know that $y=e^{\ln(y)}$. So $$\lim_{n\rightarrow \infty}(1+{1\over n})^n=\lim_{n\rightarrow \infty} y=\lim_{n\rightarrow \infty} e^{\ln(y)}=e^1=e.$$ Thus $$\lim_{n\rightarrow \infty} (1+{1\over n})^n=e.$$

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+1 for showing the correct evaluation. – Kevin Nov 4 '13 at 16:24
@Kevin :) Thank you! – 1233dfv Nov 4 '13 at 16:30

$1^{\infty}$ is an indeterminate form: the base is going to 1 which is trying to bring the expression to 1, but the exponent is going to $\infty$ which is trying to pull the expression off to $\infty$ as well. That there are these opposing forces is how I try to explain why some forms of limits are indeterminate but other forms (like $0^{\infty}=0$) aren't.

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In the expression $1^\infty $, the base is fixed. – Ittay Weiss Nov 3 '13 at 19:34
@IttayWeiss But then how do you interpret $0^0$? Shouldn't both also be fixed? – N. S. Nov 3 '13 at 21:15
$0^0$ is simply undefined. And the expression $1^\infty $ is something that is open for interpretation since the use of $\infty$ implies a limit. But in $0^0$ there is no limit implied and the answer is not uniquely determined by either the rules of algebra or the rules of common sense and applicability. – Ittay Weiss Nov 3 '13 at 21:38
@IttayWeiss In most Calculus textbooks in North America, the convention is that any expression of the type $ \lim (f(x))^{g(x)} =a^b$ simply means that $\lim f(x)=a$ and $\lim g(x)=b$. Now, if $a^b$ makes sense as a number, the continuity of $x^y$ implies that this expression, understood in this conventional sense, is the same as the number $a^b$...... Note that $1^\infty$ makes sense with this convention, but it doesn't make sense in the standard sense..... Of course people not familiar with this convention might find other meanings for it, whcih could look or be more natural..... – N. S. Nov 3 '13 at 22:17
Since $\infty $ is not a number, expressions such as $1^\infty $ are vague expressions at best and should be avoided. I am aware of the practice, especially in calculus classes, to pretend that $\infty$ is a number and use all sorts of half-baked conventions. This is a bad practice. – Ittay Weiss Nov 3 '13 at 22:25

I think it's instructive to look at what happens as n grows. At the bottom of my answer I've attached a plot showing $(1 + \frac{1}{n})^n$ as a function of n from 1 to 100.

Hopefully this plot makes a little intuitive sense. Although as $n\to\infty$, $(1 + \frac{1}{n})$ gets very small, you are raising it to a higher and higher power.

For example, when $n = 10$:

$$1 + \frac{1}{10} = 1.1$$

which is pretty close to $1$, but!

$$1.1^{10} \approx 2.6$$

e as a limit

To answer your question: You cannot substitute the limiting value in the expression and get the proper limit. As an example, consider

$$f(x) = \frac{(x^2 - 1)}{x-1}$$

If we try to take the limit as $x \to 1$ by simply substituting $x = 1$, we get

$$\lim_{x \to 1}f(x) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}$$

whereas the true limit can be found by factoring the numerator:

$$f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1$$


$$\lim_{x \to 1}f(x) = 2$$

See the Wikipedia page on limits.

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1) you probably meant to say that $1+\frac{1}{n}$ gets very close to $1$, rather than say that it gets very small. 2) The example you give later is a bit problematic for the purposes of the question. The substitution you give fails because of the term $0/0$. But then you show the correct computation, where you can eventually substitute and get the right answer. The issue is that substitution to compute limits is a good technique, but while it may work (under the suitable conditions), it never applies to substituting $\infty$. $\infty $ is not a number. – Ittay Weiss Nov 3 '13 at 21:42
@IttayWeiss, 1) Yes you are right; I just noticed that. 2) You raise a good point. I initially thought the major mistake was the substitution, but in fact it is a combination of the substitution and $\infty$. I guess the second example, although it doesn't clearly illuminate the poster's mistake, still goes to show that simple substitution does not always work. I think the method of plotting the result for increasing n is a very good way to visualize what is happening during the limiting process. – user545424 Nov 3 '13 at 21:53

This is actually a fairly good intuitive way to think about the problem but, as you point out, it does lead to the wrong answer. The mistake you're making is to use intuition instead of recognised theorems to manipulate limits. Get a textbook (or website) on limits, look up the results, and check that your assertions about the manipulations follow from these theorems. Besides your wrong answer of 1, very similar reasoning to yours can be used to obtain the answer of $\infty$. This false reasoning would work as follows. If $x > 1$, then $x^\infty = \infty$. Since $1 + \frac{1}{n}$ is always we greater than $1$, the limit is $(1 + \frac{1}{n})^\infty = \infty$ Unsurprisingly, the actual answer of $e$ lies between these two wrong answers. To actually solve your problem, evaluate the complete expression of $(1 + \frac{1}{n})^n$. Imagine that n is a large real number (like a million) and examine the expansion. You should see that it is similar to the standard expression for $e$, and that should help you to prove formally that the expansion tends to $e$ as $n$ tends to $\infty$.

Paul Epstein

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Please use proper latex command to set up your answer. – Ittay Weiss Nov 3 '13 at 19:42
@IttayWeiss Seriously? I've seen worse than that, the answer is perfectly readable as-is. – Thomas Nov 3 '13 at 21:19
@Thomas, yes, seriously. – Ittay Weiss Nov 3 '13 at 21:33

The problem is $1 + 1/n \to 1$ as $n \to \infty$, which would tend to make the limit close to or equal to $1$, but at the same time the exponent $n$ is approaching infinity, which would tend to make the limit very large. It's not apparent which is more important, and whether the limit is $1$, $\infty$, some number in between, or maybe it doesn't exist at all. It turns out that, roughly speaking, the convergence of $1 + 1/n$ to $1$ and the convergence of $n$ to $\infty$ are equally important, and the limit turns out to be something "in between", namely $e$. Someone will probably post a complete proof that the limit is $e$, but that is the idea.

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Noting : $(1+1/n)^n=e^{n \log(1+1/n)}$

And when $x\to 0$: $\log(1+x) = x+o(x)$

Or again $n \to \infty$: $log(1+1/n)=1/n+o(1/n)$

Scope : $(1+1/n)^n=e^{1+o(1)}$

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The rule of thumb with these "determinate" and "indeterminate" forms is if you can think of a way to screw it up, i.e., get a different answer depending on how you look at it, it's indeterminant. For example, $\frac{0}{0}$, the classic indeterminate form, is broken by $1 = \frac{x}{x} = \frac{0}{0}$ as $x\to 0$ and $x=\frac{x^2}{x}$ goes to $0$ as $x\to 0$. So we can equally interpret $\frac{0}{0}$ as $1$ or as $0$. Put another way, it really depends on how fast the numerator and denominator approach $0$ relative to one another.

The way to think of $1^\infty$ is this. $$x^\infty=\begin{cases}\infty,&\text{if }x > 1\\0,&\text{if }0 < x < 1\end{cases}$$ (more formally $\lim_{a\to\infty}x^a$ equals the right-hand side). So if you interpret $1^\infty$ using the first formula, you get $\infty$, and if you interpret in using the second, you get $0$. You can interpret it a third way as you saw as $1^a = 1$. What it boils down to is that $1^\infty$ really depends on how fast the base approaches one relative to how fast the exponent approaches $\infty$. You can get any number you want by varying the rate (or at least any nonnegative number), including $e$.

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Answer: It is same thing to prove $(1+x)^{1/x}$ as $x$ goes to $0$ its limit is e. Say $(1+x)^{1/x}=y$ then take log of both side. Then take limit as x approaches to 0 of both sides. Then Use L'Hopital. You will get exactly e.

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