I am trying to show that lim(n->infinity)(1+x/n)^n = e^x. A very common fact however, I am having difficulty in prooving it. (there is no proof in my textbook!).

This is how I have done it...

I took natural logarithm of the left hand side knowing that I could invert it later on!

lim(n->infinity) ln(1+x/n)^n = lim(n->inf) (ln(1+x/n))/(1/n) = 0/0

This is indetermined so I applied l'Hospital's rule...

lim(n->inf) {[1/(1+x/n)](-x/n^2)}/(-1/n^2) = x

Now take log of both sides and therefore...

lim(n->inf) ln(1+x/n)^n = x
lim(n->inf) (1+x/n)^n = e^x

I don't know if this is correct but I would really appreciate any suggestions/corrections on my analysis. Furthermore, if anyone has seen an easier way to do this please tell me!

Cheers. Ben.

That seems pretty simple; I can't find anything wrong with it, unless taking the derivative of ln(f(x)) requires the use of the theorem you've just proved, and I don't think it does.

Thankyou LaoTzu for your verification.

Another way is to first calculate the coefficients of the powers of x in the sequence and show that they converge to the corresponding term in the Taylor series of e^x around 0. Since, we know that the derivative of e^x is e^x, then we can calculate the Taylor series around x=0. It is:

e^x = \sum_{n=0}^{\infty} (x^n)/n!.

I think the kth term (ie, the term that has x^k in it) of (1 +x/m)^m is something like m chose k divided by m^k, ie:

m!/(k!(m-k)!)/m^k.

In the limit as m goes to infinity, m!/(m-k)!/m^k goes to 1, and hence the x^k coefficient goes to 1/k! which is the kth term in the Taylor series for e^x.

Excellent khallow, I would never have thought of doing it that way! Probably easier than the way I did it too!

Thanks

Looking at the original way you did it, I think also you need to handle the assumption that ln(x) is the inverse function of e^x. That shouldn't be too hard. Namely, define:

ln(x) := \integral_{1}^{x} (1/y) dy.

Then define f(x) (= e^x) as:

a) f(0) =1,

b) f'(x) = f(x).

This uniquely defines f(x), which we know happens to be e^x. The differential equation together with the definition of ln(x) will eventually (I'm handwaving here) yield ln(x) is the inverse function of e^x. Then you can take the log of (1+x/n)^n and do that stuff you did.

I think that taking natural logarithms of both sides might be assuming some of the properties you are trying to prove. The preferred way to prove this limit is to use the binomial expansion.

(1 + x/n)ⁿ = 1 + C(n,1)*(x/n)¹ + C(n,2)*(x/n)² + C(n,3)*(x/n)³ . . .

Where C(n,k) = n!/k!*(n-k)!

(1 + x/n)ⁿ = 1 + n*(x/n)¹ + (n*(n-1)/2!)*(x/n)² + (n*(n-1)*(n-2)/3!)*(x/n)³ . . .

(1 + x/n)ⁿ = 1 + x + ((n² - n)/2!)*x²)/n²*2! + ((n³ - 3n² + 2)*x³)/n³*3! . . .

As n grows without bound, the right side of the above approaches the following.

1 + x + x²/2! + x³/3! + . . . which is e.

Note that after the first two terms, each term has n^k and lower powers of n in the numerator and has n^k in the denominator.

The lower powers divided by n^k approach zero as n grows without bound, while n^k/n^k = 1 for all values of n.

I hope there are no typo's in the above.

Great feedback guys. Thanks for that.

Actually, I assumed for all real x, hence I could take logs of both sides.

Dinosaur, yet another good proof. Thankyou for the work.

Cheers!! Ben :) :)