1. I understand that the result of (1+(1/x))<sup>x</sup> as x gets very large approaches e. What I'm confused on is this... Is what I just stated not the same thing as lim x-->infinity (1+(1/x))<sup>x</sup>? I'm asking because it seems to me that the answer of lim x-->infinity (1+(1/x))<sup>x</sup> is 1. I'm obviously a bit confused. Thanks.

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3. They mean basically the same thing. What makes you believe that lim (1+(1/x) )<sup>x</sup> as x --> infinity is 1?

I'll work on a proof in the mean time if I don't get beaten to it by someone else first.

4. Actually, either I forgot something or I don't know how to do the proof. I'm thinking it might lie somewhere in my Chapter 4: Applications of Diferentiation, but I really don't see how to separate the x<sup>x</sup> into something solvable.

5. Originally Posted by Chemboy
I understand that the result of (1+(1/x))<sup>x</sup> as x gets very large approaches e. What I'm confused on is this... Is what I just stated not the same thing as lim x-->infinity (1+(1/x))<sup>x</sup>? I'm asking because it seems to me that the answer of lim x-->infinity (1+(1/x))<sup>x</sup> is 1. I'm obviously a bit confused. Thanks.
I don't think so. Are you using the intuitive notion that 1/x will tend to 0 so that you are left with (as x approaches infinity) approx 1<sup>x</sup>?

In fact you will need to think of the sum of the binomial expansion of the term (1 + (1/x))<sup>x</sup>.

This expansion contains infinite terms as x --> infinity, but each one of them can be defined/given a value. My understanding is that the sum of the series as thus expanded will tend to the value of e. Try something like Eli Maor's e: The story of a number for more info; I've left it at home so cannot quote from it but I believe it expresses exactly this formula.

cheer

shanks

6. We wish to find Z = lim x->infinity (1+1/x)<sup>x</sup>

let h = 1/x. Then:

(1+1/x)<sup>x</sup> = (1+h)<sup>1/h</sup>

Taking the natural log of this we find that:

ln((1+1/x)<sup>x</sup>) = ln((1+h)<sup>1/h</sup>) = ln(1+h)/h = [ln(1+h) - ln(1)]/h

Now, going back to the limit:

ln(Z) = ln(lim x->infinity (1+1/x)<sup>x</sup>)

By the properties of logarithms we can move the logarithm inside of the limit:

ln(Z) = lim x-> infinity ln((1+1/x)<sup>x</sup>) = lim h-> 0 ln((1+h)<sup>1/h</sup>) = lim h-> 0 [ln(1+h) - ln(1)]/h

This is just the newton quotient of ln(x) at x = 1, or:

ln(Z) = [ln(x)]'(1) = 1/1 = 1

so Z = e

Q.E.D.

7. Originally Posted by sunshinewarrio
I don't think so. Are you using the intuitive notion that 1/x will tend to 0 so that you are left with (as x approaches infinity) approx 1<sup>x</sup>?
Yeah, that's what I was doing. I kind of get the idea of what you said about the sum of the series... If this comes down to what Vroomfondel gave, it may be a bit beyond my abilities right now.

8. Chemboy: we'd love to help you understand Vroomfondel's argument. Going through the steps is a good exercise in limits, exponentials and logarithms, and derivatives. Even if you don't know calculus (specifically, derivatives), you can probably understand most of what's going on. If you know calculus, you can understand it all. So what parts don't you understand?

9. Originally Posted by serpicojr
Chemboy: we'd love to help you understand Vroomfondel's argument. Going through the steps is a good exercise in limits, exponentials and logarithms, and derivatives. Even if you don't know calculus (specifically, derivatives), you can probably understand most of what's going on. If you know calculus, you can understand it all. So what parts don't you understand?
Dunno about Chemboy, but I'd love slightly more elucidation on the last couple of steps by Vroomfondel:

lim h-> 0 [ln(1+h) - ln(1)]/h

is where my maths got me to - I couldn't have done it myself, but it looks 'right' (based on my understanding of logarithms). Then comes:

This is just the newton quotient of ln(x) at x = 1

Eh? What's a Newton quotient anyway? And then:

ln(Z) = [ln(x)]'(1) = 1/1 = 1

But but but... how did he/you make that 'h' in the denominator, that was tending to 0, disappear or cancel out?

Help!!!

10. So the "Newton quotient" is the quotient that appears in the definition of a derivative. In general, the derivative of a function f at a point x is:

f'(x) = lim<sub>h->0</sub> [f(x+h)-f(x)]/h

if the limit exists. So the expression with ln gives the derivative of ln at 1:

ln'(1) = lim<sub>h->0</sub> [ln(1+h)-ln(1)]/h

Calculus tells us that ln'(x) = 1/x, so we get ln'(1) = 1.

11. Originally Posted by serpicojr
So the "Newton quotient" is the quotient that appears in the definition of a derivative. In general, the derivative of a function f at a point x is:

f'(x) = lim<sub>h->0</sub> [f(x+h)-f(x)]/h

if the limit exists. So the expression with ln gives the derivative of ln at 1:

ln'(1) = lim<sub>h->0</sub> [ln(1+h)-ln(1)]/h

Calculus tells us that ln'(x) = 1/x, so we get ln'(1) = 1.
I knew I bloomin' hated natural logarithms nearly as much as trig. Now I remember why.

Thanks for the explanation - I just about grasped it, but it will take a lot more thinking (and possibly practice with my maths textbook at home) for the state to move from "barely grasped" to "comfortably understood".

12. I got to this in Calc. II today... What I got from what the professor said was that basically what it comes down to is that even though you can say that lim x-->infinity (1/x) = 0, it is never actually zero, and thus (1 + (1/x)) is always slightly above 1. Thus, when raised to very high powers, approaching infinity, you end up with e. Is that an ok way of looking at it? And I'm starting to get the proof that Vroomfondel gave, since I've done derivatives of ln and such in Calc. now...

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