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

[mcastillo356]

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 $n\in{\mathbb{R}}$.

 

[pasmith]

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 <br /> \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.

 

[Mark44]

Definition: Euler's Number as a Limit
(i) $e=\displaystyle\lim_{x\to{0}}{(1+x)^{\displaystyle\frac{1}{x}}}$
and
(ii) $e=\displaystyle\lim_{n\to{\infty}}{(1+\displaystyle\frac{1}{n})^n}$

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.

 

[PeroK]

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 <br /> \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]

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

 

[Mark44]

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]

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 $1$

 

[mcastillo356]

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

 

[WWGD]

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?

 

[mcastillo356]

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 $n\in\mathbb N$?.

 

[pasmith]

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 <br /> \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.

 

[Svein]

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.