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## Main Question or Discussion Point

Why does

[tex] \lim(1+\frac{1}{x})^x = e[/tex]

[tex] x->\infty[/tex]

[tex] \lim(1+\frac{1}{x})^x = e[/tex]

[tex] x->\infty[/tex]

- Thread starter johndoe
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Why does

[tex] \lim(1+\frac{1}{x})^x = e[/tex]

[tex] x->\infty[/tex]

[tex] \lim(1+\frac{1}{x})^x = e[/tex]

[tex] x->\infty[/tex]

- 586

- 1

This is one possible **definition** of the number e.

What is your definition of e?

What is your definition of e?

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This is one possibledefinitionof the number e.

What is your definition of e?

O so how do i prove this ?

rock.freak667

Homework Helper

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[tex]L=\lim_{x\rightarrow \infty} (1+\frac{1}{x})^x[/tex]

[tex]ln L =ln \lim_{x\rightarrow \infty} (1+\frac{1}{x})^x[/tex]

[tex]ln L =\lim_{x\rightarrow \infty} xln (1+\frac{1}{x})[/tex]

Then use L'Hopital's Rule twice.

Hurkyl

Staff Emeritus

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If you don't have a definition ofO so how do i prove this ?

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How do you prove what?O so how do i prove this ?

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yea how do you prove a definition? maybe he means how do you consistency?How do you prove what?

HallsofIvy

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Homework Helper

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Unfortunately, johndoe hasn't gotten back to us to answer the question about what definition of e he is using.

Yes, many texts**define** e by that limit. If that is the definition, then no "proof" is required.

Others, however, treat the derivative of f(x)= a^{x} this way:

[tex]f(x+ h)= a^{x+ h}= a^x a^h[/tex]

[tex]f(x+ h)- f(x)= a^xa^h- a^x= a^x(a^h- 1)[/tex]

[tex]\frac{f(x+h)- f(x)}{h}= a^x\frac{a^h- 1}{h}[/tex]

Which, after you have shown that

[tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex]

**exists**, shows that the derivative of a^{x} is just a constant times a^{x}.

"e" is then defined as the value of a such that that constant is 1:

[tex]\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1[/tex]

We can then argue (roughly, but it can be made rigorous) that if, for h close to 0, (e^{h}-1)/h is close to 1, e^{h}- 1 is close to h and so e^{h} is close to h+ 1. Then, finally, e is close to (h+1)^{1/h}. As h goes to 0, 1/h goes to infinity. letting x= 1/h, (h+1)^{1/h} becomes (1/x+ 1)^{x} so

[tex]\lim_{x\rightarrow \infty}(1+ \frac{1}{x})^x= e[/tex]

Here's a third definition of "e":

It has become more common in Calculus texts to work "the other way". That is, define ln(x) by

[tex]ln(x)= \int_1^x \frac{1}{t}dt[/itex]

From that we can prove all the properties of ln(x) including the fact that it is a "one-to-one" and "onto" function from the set of positive real numbers to the set of all real numbers- and so has an inverse function. Define "exp(x)" to be the inverse function . Then we define "e" to be exp(1) (one can show that exp(x) is, in fact, e to the x power).

Now, suppose [itex]lim_{x\rightarrow \infty} (1+ 1/x)^x[/itex] equals some number a. Taking the logarithm, x ln(1+ 1/x)must have limit ln(a). Letting y= 1/x, that means that the limit, as y goes to 0, of ln(1+y)/y must be ln(a). But that is of the form "0/0" so we can apply L'hopital's rule: the limit is the same as the limit of 1/(y+1) which is obviously 1. That is, we must have ln(a)= 1 or a= e.

Yes, many texts

Others, however, treat the derivative of f(x)= a

[tex]f(x+ h)= a^{x+ h}= a^x a^h[/tex]

[tex]f(x+ h)- f(x)= a^xa^h- a^x= a^x(a^h- 1)[/tex]

[tex]\frac{f(x+h)- f(x)}{h}= a^x\frac{a^h- 1}{h}[/tex]

Which, after you have shown that

[tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex]

"e" is then defined as the value of a such that that constant is 1:

[tex]\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1[/tex]

We can then argue (roughly, but it can be made rigorous) that if, for h close to 0, (e

[tex]\lim_{x\rightarrow \infty}(1+ \frac{1}{x})^x= e[/tex]

Here's a third definition of "e":

It has become more common in Calculus texts to work "the other way". That is, define ln(x) by

[tex]ln(x)= \int_1^x \frac{1}{t}dt[/itex]

From that we can prove all the properties of ln(x) including the fact that it is a "one-to-one" and "onto" function from the set of positive real numbers to the set of all real numbers- and so has an inverse function. Define "exp(x)" to be the inverse function . Then we define "e" to be exp(1) (one can show that exp(x) is, in fact, e to the x power).

Now, suppose [itex]lim_{x\rightarrow \infty} (1+ 1/x)^x[/itex] equals some number a. Taking the logarithm, x ln(1+ 1/x)must have limit ln(a). Letting y= 1/x, that means that the limit, as y goes to 0, of ln(1+y)/y must be ln(a). But that is of the form "0/0" so we can apply L'hopital's rule: the limit is the same as the limit of 1/(y+1) which is obviously 1. That is, we must have ln(a)= 1 or a= e.

Last edited by a moderator:

- 39

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Unfortunately, johndoe hasn't gotten back to us to answer the question about what definition of e he is using.

Yes, many textsdefinee by that limit. If that is the definition, then no "proof" is required.

Others, however, treat the derivative of f(x)= a^{x}this way:

[tex]f(x+ h)= a^{x+ h}= a^x a^h[/tex]

[tex]f(x+ h)- f(x)= a^xa^h- a^x= a^x(a^h- 1)[/tex]

[tex]\frac{f(x+h)- f(x)}{h}= a^x\frac{a^h- 1}{h}[/tex]

Which, after you have shown that

[tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex]

exists, shows that the derivative of a^{x}is just a constant times a^{x}.

"e" is then defined as the value of a such that that constant is 1:

[tex]\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1[/tex]

We can then argue (roughly, but it can be made rigorous) that if, for h close to 0, (e^{h}-1)/h is close to 1, e^{h}- 1 is close to h and so e^{h}is close to h+ 1. Then, finally, e is close to (h+1)^{1/h}. As h goes to 0, 1/h goes to infinity. letting x= 1/h, (h+1)^{1/h}becomes (1/x+ 1)^{x}so

[tex]\lim_{x\rightarrow \infty}(1+ \frac{1}{x})^x= e[/tex]

Here's a third definition of "e":

It has become more common in Calculus texts to work "the other way". That is, define ln(x) by

[tex]ln(x)= \int_1^x \frac{1}{t}dt[/itex]

From that we can prove all the properties of ln(x) including the fact that it is a "one-to-one" and "onto" function from the set of positive real numbers to the set of all real numbers- and so has an inverse function. Define "exp(x)" to be the inverse function . Then we define "e" to be exp(1) (one can show that exp(x) is, in fact, e to the x power).

Now, suppose [itex]lim_{x\rightarrow \infty} (1+ 1/x)^x[/itex] equals some number a. Taking the logarithm, x ln(1+ 1/x)must have limit ln(a). Letting y= 1/x, that means that the limit, as y goes to 0, of ln(1+y)/y must be ln(a). But that is of the form "0/0" so we can apply L'hopital's rule: the limit is the same as the limit of 1/(y+1) which is obviously 1. That is, we must have ln(a)= 1 or a= e.

Ok I see. I didn't know that definition before until I reach a problem requiring me to find that limit.

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e is such that when: y=e^x

dy/dx = e^x

Hence via the Taylor Series, the actual value of e can be found

right???

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