Math Is Fun Forum

Onyx

Member

Registered: 2009-02-24

Posts: 48

how does e=lim_{x->0}(1+x)^{1/x}?

Hi can someone please show me how this limit exists? I haven't been able to find anything useful online for it.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: how does e=lim_{x->0}(1+x)^{1/x}?

How much mathematics have you had?  Are you taking a calculus class or are you in analysis?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LampShade

Member

Registered: 2009-02-22

Posts: 23

Re: how does e=lim_{x->0}(1+x)^{1/x}?

The limit you referred to is 1.  I think you were actually inquiring about this:

That's just one way of defining e.  The trick isn't to prove that the limit as x goes to infinity is e; rather, it is to show that this limit exists and agrees with the other definitions that e possesses.

Check out http://web01.shu.edu/projects/reals/numseq/s_euler.html.

Last edited by LampShade (2009-02-26 11:19:38)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: how does e=lim_{x->0}(1+x)^{1/x}?

That's just one way of defining e.  The trick isn't to prove that the limit as x goes to infinity is e; rather, it is to show that this limit exists and agrees with the other definitions that e possesses.

This is entirely correct, I just wanted to be a bit more careful with the wording.  To prove that the limit of something is e, first you need to know what e is.  One way to define e is in the limit above.  But, if for example you define e as:

[align=center]

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Once you make this definition, it would now be required to prove that:

[align=center]

[/align]

Which as I was alluding to earlier, takes higher mathematics (analysis).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Onyx

Member

Registered: 2009-02-24

Posts: 48

Re: how does e=lim_{x->0}(1+x)^{1/x}?

Hi, I think there is two similar limits that equal e. There is one that is used to describe compound interest:

and there is another one, which I got from this link http://www.mathcentre.ac.uk/students.php/all_subjects/differentiation/first_principles/resources/322 which is:

I don't think you can simply substitute t=0 to get the limit as 1, since the denominator of the exponent becomes zero, so...

...is undefined. If you put in values closer and closer to 0 for x, then you will see the limit gets closer and closer to e.

Ricky, to answer your question, I'm an engineering student, and while I've done calculus and multivariable calculus before, I've forgotten how to differentiate the exponential function from first principles (I never need to do this, it's just a matter of interest), and have never really covered analysis in much depth. I know there are multiple definitions for the exponential function, and I've always considered "the function whose value equals its derivative for all x" to be the 'main' definition, so if you could show me how this limit is equivalent to that or the other limit I gave I would appreciate it.

Also I remember a derivation where the series (Taylor series?) you mentioned can be derived purely from knowing that:

and:

Where somehow my lecturer obtained a series for the function and repeatedly differentiated it. If you know what I'm talking about could you show me that too?

Thanks alot

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: how does e=lim_{x->0}(1+x)^{1/x}?

The Mclaurin series (Taylor series about x = 0) for exponentinal function is:

which ofcourse you can work backwards from:

Mclaurin series is formed by:

if:

then;

Last edited by luca-deltodesco (2009-02-26 06:35:03)

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LampShade

Member

Registered: 2009-02-22

Posts: 23

Re: how does e=lim_{x->0}(1+x)^{1/x}?

Sorry, when I first saw your question, I thought you were taking the limit as x goes to infinity for some reason.  Notice that taking x to infinity in your definition of e is the same is taking x to zero in mine, in which both cases we get e° = 1, as expected.

However, using

seems a bit more intuitive that e° = 1.

Another interesting fact is that

Last edited by LampShade (2009-02-26 11:25:39)

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LampShade

Member

Registered: 2009-02-22

Posts: 23

Re: how does e=lim_{x->0}(1+x)^{1/x}?

Last edited by LampShade (2009-02-26 18:54:33)

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