Derive the limit of an expression (1+1/n)^n

In summary, the limit of the expression is e and it can be derived by showing that it converges to a unique number, which we define as e. The definition of e is that the derivative of ex is just ex, and we can approximate e by solving for it in a series. The series converges quickly to 2.718... which is e.
  • #1
rahl__
10
0
i know that that the limit of this expression is e, but i don't know how to derive it...
i will be really grateful if someone could help me solve this problem
 
Mathematics news on Phys.org
  • #2
You would need a definition of e other than that. Otherwise, you can just show that the limit converge and call it e.

So what definition of e do you use?
 
  • #3
If a is any positive number then
[tex]lim_{h\rightarrow0}\frac{a^{x+h}-a^x}{h}= lim_{h\rightarrow0}\frac{a^xa^h- a^x}{h}[/tex]
[tex]= a^x lim{h\rightarrow0} \frac{a^h- 1}{h}[/tex]

In other words, ax has the nice property that its derivative is just a number (that limit) time ax itself.

We define e to be the number such that the derivative of ex is just ex- in other words so that that
[tex]lim_{x\rightarrow0}\frac{e^h-1}{h}= 1[/tex].

That means that for small h,
[tex]\frac{e^h-1}{h}[/tex]
is approximately 1.

Let n= 1/h (more accurately the next integer larger than 1/h). Then
[tex]n\left(e^{\frac{1}{n}}-1\right)= 1[/tex]
approximately for large n with 1 being the limit as n goes to infinity.

Solve that for e: e is approximately
[tex]\left(1- \frac{1}{n}\right)^n[/tex]
with the limit being e as n goes to infinity.
 
  • #4
i want to show that 2,718... [which is e] is the limit converge of that expression, I've posted my question in this 'strange' way to avoid getting answers such as "under the definition of e the limit converge of that expression is e".
hope u understand what I am talking about

Otherwise, you can just show that the limit converge and call it e.
thats what i trying to ask about
 
Last edited:
  • #5
Well, you can always proceed [tex]limn\rightarrow\infty(1+\frac{1}{n})^n=1+n(1/n)+(n)(n-1)(1/n^2)1/2!+n(n-1)(n-2)(1/n^3)1/3!+++[/tex]

Notice here that the numerator of each term we have the power of n that corresponds to the power in the demoninator, other than that, the numenator has lower powers of n and these can be eliminated since

[tex]limn\rightarrow\infty\frac{n-1}{n}=(n/n-1/n)\rightarrow1[/tex] (Here we have to consider that, for example, [tex]\frac{n(n-1)(n-2)}{n^33!}\leq1/3![/tex] We proceed to do this with every term in the series.

This leaves us with the series 1+1+1/2!+1/3!+1/4!+++

Now as my professor once said, e gets to its limit very fast. He added on the blackboard 1+1+1/2+1/6+1/24 +1/120+1/720+1/5040+1/43202=2.718.. (He did this all using decimals way before calculators.)
 
Last edited:
  • #6
got it, thanks
 

FAQ: Derive the limit of an expression (1+1/n)^n

What is the limit of the expression (1+1/n)^n as n approaches infinity?

The limit of the expression (1+1/n)^n as n approaches infinity is equal to the mathematical constant e, also known as Euler's number, which is approximately 2.71828.

How do you derive the limit of (1+1/n)^n?

To derive the limit of (1+1/n)^n, we can use the fact that as n approaches infinity, the expression (1+1/n) approaches 1. Therefore, we can rewrite the expression as (1+1/n)^n = (1)^n = 1. This shows that the limit is equal to 1.

Can you explain why the limit of (1+1/n)^n is equal to e?

The limit of (1+1/n)^n is equal to e because as n approaches infinity, the expression becomes closer and closer to the definition of e, which is the limit of (1+1/n)^n as n approaches infinity. This can be seen through the graph of the function y = (1+1/x)^x, which approaches the graph of y = e as x approaches infinity.

Is it possible for the limit of (1+1/n)^n to be any other value besides e?

No, the limit of (1+1/n)^n can only be equal to e because it is a special case of the more general limit of (1+1/n)^k as n approaches infinity, where k is any constant. This limit is equal to e for all values of k, as shown by the graph of the function y = (1+1/n)^k, which approaches the graph of y = e as n approaches infinity for all values of k.

What is the significance of the limit of (1+1/n)^n?

The limit of (1+1/n)^n, which is equal to e, has many important applications in mathematics, physics, and other sciences. It is used to calculate compound interest, population growth, and radioactive decay, among other things. It also plays a key role in the definition of the natural logarithm, and has connections to complex numbers and fractals.

Similar threads

Replies
14
Views
2K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
7
Views
1K
Replies
5
Views
1K
Back
Top