15

I am trying to identify what the flaw is exactly when reasoning about a limit such as the definition of $\mathbf e$:

$$ \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n={e} $$

Now, I know there are ways of proving this limit, such as by considering the binomial expansion of $(1+\frac{1}{n})^n$ and comparing that to the Maclaurin series of $e$. So to make this clear, I am not looking for a proof of the limit definition of $e$.

I tried searching for "limit laws/rules" but none of the rules I found described the above case. Hence, I am looking for a specific rule (or perhaps a certain perspective) that will help me realize why the above does not in fact evaluate to 1.

My train of thought is as follows: at a first glance, if I wasn't already familiar with what the limit evaluates to, I probably would evaluate the expression inside the brackets first, and then apply the limit to the power. So, $$\displaystyle\lim_{n\to\infty}\frac{1}{n}=0\quad\text{and then}$$ $$\displaystyle\lim_{n\rightarrow \infty}\left(1+0\right)^n=\lim_{n\to\infty}1^n=1$$

Why is my reasoning flawed?

Martin Sleziak
  • 53,687
Milo Wielondek
  • 1,919
  • 14
    this may help. – David Mitra Apr 25 '12 at 13:27
  • 15
    Sometimes it helps to look at a simpler version of the flawed reasoning, with extraneous details removed. So consider $\lim_{n\to\infty}n\cdot\frac{1}{n}$. $\lim_{n\to\infty}\frac{1}{n}=0$, and so $\lim_{n\to\infty}n\cdot\frac{1}{n}=\lim_{n\to\infty}n\cdot 0=\lim_{n\to\infty} 0=0$. – Chris Eagle Apr 25 '12 at 13:30
  • You could also try taking logarithm (it is a continuous function, so the limits should preserve), and arrive at $\lim_{n\to\infty}n\log(1+1/n)$ what is of the form $\infty\cdot 0$. – dtldarek Apr 25 '12 at 13:32
  • 1
    Related to @ChrisEagle's answer, $e^{1/n}\rightarrow 1$ as $n\rightarrow \infty$. Your argument would say that $e=(e^{1/n})^n\rightarrow 1$. This is close to what is happening here - $1+\frac{1}{n}$ is a "good" approximation of $e^{1/n}$ when $n$ is large. – Thomas Andrews Apr 25 '12 at 13:44
  • 8
    You are applying the "rule" $\lim f(n)^{g(n)}=\bigl(\lim f(n)\bigr)^{\lim g(n)}$ which is unjustified. – Andrea Mori Apr 25 '12 at 13:47
  • 8
    There is a widespread feeling that deciding whether $\lim_{x\to a}f(x)$ exists, and computing the limit when it exists, is the same thing as "evaluating" $f$ at $a$. There is also a widespread feeling that "$\infty$" is a number, like $123456$ but bigger. Neither is the case. But the feelings combine to lead to the belief that $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ is an expression much like $\left(1+\frac{1}{5}\right)^5$, and subject to the same evaluation rules. – André Nicolas Apr 25 '12 at 14:17
  • 5
    $n=n$. You can't let one $n$ go to infinity without the other! What you compute (correctly) is $\lim_n (\lim_m (1+1/m)^n)$. – Matthew Towers Apr 25 '12 at 15:22
  • @AndréNicolas since I too share these feelings, where can I turn to read up on what 'the case' actually is? – Milo Wielondek May 03 '12 at 22:49
  • 1
    @Milosz: I assume you are taking a course about these things now. It takes a while to develop intuition. Making mistakes is part of the process. – André Nicolas May 03 '12 at 22:59

6 Answers6

16

You can see the same problem with something like $n^{1/\log n}$. If you think "of $n$ first", then the limit would be infinite. If you think "of $1/\log n$ first", the limit would be $1$. But it's neither! (as Chris mentioned above, this already happens with products, and even with sums).

There is no justification in isolating one part from the other: each new value of $n$ obviously implies a new value of $1/\log n$, so there's no ground in thinking that they should behave "independently".

This mistake is probably induced by the fact that there is actually "independence" in several situations, like "the limit of the sum is the sum of the limit", or "the limit of the product is the product of the limits"; but even these situations require both limits to exist. The power situation is no different, in the sense that you cannot freely "distribute" the limit.

Martin Argerami
  • 205,756
  • 6
    It's good, as you suggest, to think of the base $n$ and the exponent $1/n$ as contributing opposite effects as $n \rightarrow \infty$. We have a big base to a small power; it's not clear what we get unless we know just how big and how small the numbers are! But in your example, the effect of the exponent is strong enough to fully determine the answer; a very small exponent will give you an answer close to 1, and the limit of $n^{1/n}$ is in fact 1. But $n^{1/\log(\log n)}$ goes to infinity (the effect of the base dominates); and $n^{1/\log n}$ goes to $e$ (a compromise between the two). – Jonas Kibelbek Apr 25 '12 at 15:16
10

A similar flaw occurs in this reasoning:

What is $\displaystyle\lim_{n\to0}\left(n\cdot\frac1n\right)$? Well, evaluating the first term first, we have: \begin{align} \lim_{n\to0}\left(n\cdot\frac1n\right)&=\lim_{n\to0}\left(0\cdot\frac1n\right)\\ &=\lim_{n\to0}0\\ &=0 \end{align}

This argument is clearly in error, as $\lim_{n\to0}\left(n\cdot\frac1n\right)$ is equal to $\lim_{n\to0}1=1$.

The flaw with this argument is that you can't evaluate a limit part-by-part. You need to consider the thing as a whole. In fact, since $\lim_{n\to0}n=0$ and $\lim_{n\to0}\frac1n=\infty$, the above limit is of the form $0\cdot\infty$. This is called an indeterminate form: Just knowing that a limit is of the form $0\cdot\infty$ doesn't tell you anything about the value of the limit.

Your example, $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n$, is also an indeterminate form. In this case, we have a limit of the form $1^\infty$. Since it's an indeterminate form, we can't evaluate it part-by-part, only as a whole.

A list of indeterminate forms: $\infty-\infty$, $\dfrac00$, $\dfrac\infty\infty$, $0\cdot\infty$, $1^\infty$, $\infty^0$, $0^0$.

(The last one is odd in that it's hard to find examples that aren't equal to $1$, other than the easy $\lim_{x\to0^+}0^x$. There's a reason for this that I won't get into. An example where it's not equal to one: $\displaystyle\lim_{x\to0}\left(e^{-1/x^2}\right){}^{x^2}=\frac1e$.)

Akiva Weinberger
  • 23,062
5

The limit must clearly exceed $1$. Here is why.

Take $$\left(1+\frac1n\right)^n$$and expand it using the binomial formula.

For any $n>0$, you get a sum of positive terms, the first two being $1$:

$$\color{green}{1+\frac nn}+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}+\cdots\frac1{n^n}\color{green}{\ge2}.$$

When you increase $n$, every $k^{th}$ tends to $\dfrac1{k!}$ because $k$ becomes neglectible in front of $n$, leading to the well-known sum

$$1+1+\frac12+\frac1{3!}+\frac1{4!}+\cdots.$$

1

There is a better way of approaching when the indeterminate form is either $1^\infty$ or $0^0$.

Write the given function as $e^{log(f(x))}$.

There is a rule for $\displaystyle\lim_{n\to a}f(x)^{g(x)}=\displaystyle\lim_{n\to a}f(x)^{\displaystyle\lim_{n\to a}g(x)}$.

So you can apply L'Hospital's rule to evaluate the limit in the given case which comes out to be e.

Vinanth S Bharadwaj
  • 420
1

As noticed, the key point is that for limits in the form $f(x)^{g(x)}$ with $f(x)\to 1$ and $g(x)\to \infty$, by substitution or solving the limit part-by-part, we obtain $1^\infty$ which is an indeterminate form that can lead to different answers depending on the particular nature of $f(x)$ and $g(x)$.

To see why this is an indeterminate form, let consider $$f(x)^{g(x)}=e^{g(x)\log(f(x))}$$

with $g(x)\log(f(x))$ in the more intuitive indeterminate form $\infty \cdot 0$.

Note also that the case $f(x)=1$ doesn't lead to an indeterminate form, indeed $\forall g(x)$

$$f(x)^{g(x)}=1^{g(x)}=1$$

and also, more in general, the case $f(x)\to L\in \mathbb R^+\setminus\{1\}$ doesn't leads to an indeterminate form, and we can proceed by subsitution as follows

$$f(x)^{g(x)}\to L^\infty\begin{cases}=\infty \;\text{for} \;L>1\\=0\;\text{for} \;0<L<1\end{cases}$$

It is important to observe that in such cases we can proceed by a first evaluation solving the limit part-by-part as in the following examples

$$\lim_{n\rightarrow \infty}\left(2+\frac{1}{n}\right)^n=(2+0)^\infty =\infty$$

$$\lim_{n\rightarrow \infty}\left(\frac12+\frac{1}{n}\right)^n=\left(\frac 12+0\right)^\infty =\left(\frac 12\right)^\infty=0$$

which leads to correct results.

user
  • 154,566
1

Bernouilli's inequality:

$(1+1/n)^n \ge 1+n(1/n) =2$ for $ n=1,2,3...$

$2$ is a lower bound.

Peter Szilas
  • 20,344
  • 2
  • 17
  • 28