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

1. Mar 6, 2006

### rahl__

i know that that the limit of this expression is e, but i dont know how to derive it...
i will be really grateful if someone could help me solve this problem

2. Mar 6, 2006

### quasar987

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. Mar 6, 2006

### HallsofIvy

If a is any positive number then
$$lim_{h\rightarrow0}\frac{a^{x+h}-a^x}{h}= lim_{h\rightarrow0}\frac{a^xa^h- a^x}{h}$$
$$= a^x lim{h\rightarrow0} \frac{a^h- 1}{h}$$

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
$$lim_{x\rightarrow0}\frac{e^h-1}{h}= 1$$.

That means that for small h,
$$\frac{e^h-1}{h}$$
is approximately 1.

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

Solve that for e: e is approximately
$$\left(1- \frac{1}{n}\right)^n$$
with the limit being e as n goes to infinity.

4. Mar 6, 2006

### rahl__

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 im talking about

Last edited: Mar 6, 2006
5. Mar 6, 2006

### robert Ihnot

Well, you can always proceed $$limn\rightarrow\infty(1+\frac{1}{n})^n=1+n(1/n)+(n)(n-1)(1/n^2)1/2!+n(n-1)(n-2)(1/n^3)1/3!+++$$

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

$$limn\rightarrow\infty\frac{n-1}{n}=(n/n-1/n)\rightarrow1$$ (Here we have to consider that, for example, $$\frac{n(n-1)(n-2)}{n^33!}\leq1/3!$$ 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: Mar 6, 2006
6. Mar 6, 2006

### rahl__

got it, thanks