# Naturals or Reals When Taking Limits to Obtain the Value of Euler's Number e?

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TL;DR Summary
I've found two ways of taking limits in order to obtain ##e## number. Do they belong to naturals in one case and to the reals in the second case?
Hi PF

Searching on the Internet, I've found this definition:

Definition: Euler's Number as a Limit

(i) ##e=\displaystyle\lim_{x\to{0}}{(1+x)^{\displaystyle\frac{1}{x}}}##

and

(ii) ##e=\displaystyle\lim_{n\to{\infty}}{(1+\displaystyle\frac{1}{n})^n}##

Questions:
1-Does it make sense Definition (i)? I don't think so: ##(1+x)## on the base is strange: a lonely ##x## on the sum.
2-If both make sense, does ##x\in{\mathbb{R}}## in (i), and ##n\in{\mathbb{N}}## in (ii)?.

Attempt: I think I've fallen into an erratic web; but let's suppose the contrary. In that case, no matters reals neither naturals. I make no distinction. It could be also ##x\in{\mathbb{N}}##, and ##n\in{\mathbb{R}}##.

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We have the following general proposition:

If $\lim_{x \to c} f(x)$ exists and $(x_n)$ is a real sequence such that $\lim_{n \to \infty} x_n = c$ then $$\lim_{n \to \infty} f(x_n) = \lim_{x \to c} f(x).$$ This holds also for one-sided limits, provided $x_n \to c$ from the appropriate direction.

vanhees71, mcastillo356 and PeroK
Mentor
Definition: Euler's Number as a Limit
(i) ##e=\displaystyle\lim_{x\to{0}}{(1+x)^{\displaystyle\frac{1}{x}}}##
and
(ii) ##e=\displaystyle\lim_{n\to{\infty}}{(1+\displaystyle\frac{1}{n})^n}##
mcastillo356 said:
1-Does it make sense Definition (i)? I don't think so: ##(1+x)## on the base is strange: a lonely ##x## on the sum.
2-If both make sense, does ##x\in{\mathbb{R}}## in (i), and ##n\in{\mathbb{N}}## in (ii)?.
1. Both are equally valid definitions of the number ##e##. Any positive real number not equal to 1 can be a base of an exponential expression. A negative base results in problems with continuity.

I don't understand what "a lonely ##x## on the sum" means.

2. Yes. For the first definition, x is a real number "close to" 1. In the second, n is an integer.

mcastillo356
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Gold Member
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We have the following general proposition:

If $\lim_{x \to c} f(x)$ exists and $(x_n)$ is a real sequence such that $\lim_{n \to \infty} x_n = c$ then $$\lim_{n \to \infty} f(x_n) = \lim_{x \to c} f(x).$$ This holds also for one-sided limits, provided $x_n \to c$ from the appropriate direction.
In fact, an alternative, equivalent definition of a limit uses sequences:

The function ##f## has limit ##L## at the point ##a## iff for all sequences ##\{x_n\} \in D - \{a\}## (where ##D## is the domain of ##f##) with ##\lim_{n \to \infty} x_n = a## we have ##\lim_{n \to \infty} f(x_n) = L##.

This can be shown to be equivalent to the more usual epsilon-delta definition. I think it's enormously useful, especially if you want to prove that a limit does not exist.

mcastillo356
Gold Member
1. Both are equally valid definitions of the number ##e##.
Perfect.
Any positive real number not equal to 1 can be a base of an exponential expression. A negative base results in problems with continuity.

I don't understand what "a lonely ##x## on the sum" means.

2. Yes. For the first definition, x is a real number "close to" 1. In the second, n is an integer.
What means the first sentence? ##1^5## is not an exponential expression? Why can't ##1## be a base? Is it because is iterative? I mean, no matter the exponent, always equals ##1##.
"a lonely ##x## on the sum" is an unfortunate thought. I refer to the first quote (#1)
In the second, n is an integer is mentioned. Sequences aren't meant to deal with naturals?

Working on #2 and #4 post. Not sure to achieve

Mentor
What means the first sentence? ##1^5## is not an exponential expression? Why can't ##1## be a base? Is it because is iterative? I mean, no matter the exponent, always equals ##1##.
Yes, ##1^5## is an exponential expression, but as you note, ##1^5## is just a convoluted way to write 1. What I meant was that when we talk about the base of an exponential expression, we usually exclude 1 as a possible base because ##1^x## isn't very interesting. All of the other exponential functions, where the base is positive but not equal to 1, have inverses -- ##\log## functions.
Sequences aren't meant to deal with naturals?
Yes, a sequence can be defined as a function whose domain is the natural numbers, but the 2nd definition you showed is a limit, not a sequence. You could turn it into a sequence by the substitution ##n = \frac 1 x##, where x is chosen so that its reciprocal is an integer.

mcastillo356
Gold Member
Yes, ##1^5## is an exponential expression, but as you note, ##1^5## is just a convoluted way to write 1. What I meant was that when we talk about the base of an exponential expression, we usually exclude 1 as a possible base because ##1^x## isn't very interesting. All of the other exponential functions, where the base is positive but not equal to 1, have inverses -- ##\log## functions.
Absolutely sound paragraph. Number ##1## is tricky, slippery, and a ground for brainy mathematicians.

Yes, a sequence can be defined as a function whose domain is the natural numbers, but the 2nd definition you showed is a limit, not a sequence. You could turn it into a sequence by the substitution ##n = \frac 1 x##, where x is chosen so that its reciprocal is an integer.
Brilliant! Good argument, helpful to judge, decide about, and eventually deal with number ##1##

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Hi, PF, I'm working on second and fourth posts of the thread. Second is very interesting, and #4 seems to be an alternative proposal; that is, they state the same argument.
Next I write my point of view, after a quickly searching. Actually, is a copy and paste. Finally, I'll ask the doubt. Here it goes:
Limit and sequences
We consider the set of all sequences ##\{x_n\}## such that

a) ##x_n\neq{a}\;\forall{n}##
b) ##\lim_{n\to\infty}{x_n=a}##

The fofllowing characterization of existence of limit verifies:
A function ##f## has got limit ##l## at a point ##a## iff for any sequence ##x_n## satisfying a) and b) it verifies ##\lim_{n\to\infty}{f(x_n)=l}##
The previous condition is useful to prove that some functions do not have limit at a point ##a##; it is enough to find a sequence ##x_n## which elements are different to ##a##, such that ##\lim_{n\to\infty}{x_n=a}##, and ##\lim_{n\to\infty}{f(x_n)}## doesn't exist. Other way to use the previous criterion in the opposite sense is to find two different sequences ##x_n## and ##y_n##, verifying a) and b) properties, such that
##\lim_{n\to\infty}{f(x_n)}\neq{\lim_{n\to\infty}{f(y_n)}}##

Example
Consider ##f(x)=\sin{(1/x)}## for ##x\neq{0}##. Prove it has no limit at ##x=0##, making use of sequences:

Take ##x_n## and ##y_n##, convergent to ##0##, such that
##x_n=\displaystyle\frac{1}{2\pi n}##
and
##y_n=\displaystyle\frac{1}{\dfrac{\pi}{2}+2\pi n}##, ##n\in{\mathbb{N}}##
As ##f(x_n)=\sin{\Big(\dfrac{1}{x_n}\Big)}=\sin{(2\pi n)}=0\;\forall{n}##, we have that ##\lim_{n\to\infty}{f(x_n)}=0##
On the other hand, ##f(y_n)=\sin{\Big(\dfrac{1}{y_n}\Big)}=\sin{\Big(\dfrac{\pi}{2}+2\pi n\Big)}=1\;\forall{n}##; therefore ##\lim_{n\to\infty}{f(y_n)}=1##.

This proves that function ##f##, whose graph is shown, has got no limit at ##0##.

Doubt
Is this speech near to stated at 2nd and fourth posts?

Sorry, bad LaTeX. Hope is understandable

Last edited:
Gold Member
Re
TL;DR Summary: I've found two ways of taking limits in order to obtain ##e## number. Do they belong to naturals in one case and to the reals in the second case?

Hi PF

Searching on the Internet, I've found this definition:

Definition: Euler's Number as a Limit

(i) ##e=\displaystyle\lim_{x\to{0}}{(1+x)^{\displaystyle\frac{1}{x}}}##

and

(ii) ##e=\displaystyle\lim_{n\to{\infty}}{(1+\displaystyle\frac{1}{n})^n}##

Questions:
1-Does it make sense Definition (i)? I don't think so: ##(1+x)## on the base is strange: a lonely ##x## on the sum.
2-If both make sense, does ##x\in{\mathbb{R}}## in (i), and ##n\in{\mathbb{N}}## in (ii)?.

Attempt: I think I've fallen into an erratic web; but let's suppose the contrary. In that case, no matters reals neither naturals. I make no distinction. It could be also ##x\in{\mathbb{N}}##, and ##n\in{\mathbb{R}}##.
Remember that naturals are Real numbers . Idea in both can be seen as compounding a very small number infinitely -many times. So one , the 1+1/x will be "pulling " to remain close to 0, while the exponent will "pull" in the opposite direction.
To verify these are equivalent, try the change of variable 1/n =x .

Is that what you were asking?

Gold Member
Re

Remember that naturals are Real numbers . Idea in both can be seen as compounding a very small number infinitely -many times. So one , the 1+1/x will be "pulling " to remain close to 0, while the exponent will "pull" in the opposite direction.
To verify these are equivalent, try the change of variable 1/n =x .

Is that what you were asking?
Very enriching words. But I was asking the Forum to check my point of view about these two limits to obtain ##e##, which I consider as a ##\lim_{x\rightarrow{\infty}}##, for ##x\in{\mathbb{R}}##, on one side, and the equivalent way to reach ##e##, through the limit of a sequence, ##\Big(1+\frac{1}{n}\Big)^n##, on the other side; this second approach is made within ##n\in\mathbb N##?.

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Very enriching words. But I was asking the Forum to check my point of view about these two limits to obtain ##e##, which I consider as a ##\lim_{x\rightarrow{\infty}}##, for ##x\in{\mathbb{R}}##, on one side, and the equivalent way to reach ##e##, through the limit of a sequence, ##\Big(1+\frac{1}{n}\Big)^n##, on the other side; this second approach is made within ##n\in\mathbb N##?.

The limit $$\lim_{x \to \infty} (1 + x^{-1})^x$$ can also be taken through real values of $x$, and a proposition cited earlier then leads to $\lim_{n \to \infty} (1 + x_n^{-1})^{x_n} = e$ for any sequence $x_n \to \infty$, and one is free to choose $x_n = n$.

PeroK and mcastillo356
In general, if a countable sequence of function values converge and has a limit, it does not imply that the function itself has a limit. Example: $\lim_{n\rightarrow \infty}( \sin(n\cdot\pi)) = 0$ since all elements in the sequence are 0, but $\lim_{x\rightarrow \infty}( \sin(x))$ does not exist.