# Napier's Constant Limit Definition

1. Mar 31, 2012

### scorsesse

Hi all ! I am terribly sorry if this was answered before but i couldn't find the post. So that's the deal. We all know that while x→∞ (1+1/x)^x → e

But I am deeply telling myself that 1/x goes to 0 while x goes to infinity. 1+0 = 1 and we have 1^∞ which is undefined. But also see that 1/x +1 is not a continuous function so i cannot simply take the limit of it and raise the value to x like : (limit of 1/x + 1)^x while x→∞

So can you please give me a rigorous proof for why this function approaches to Napier's constant ?

2. Mar 31, 2012

### HallsofIvy

You cannot first take the limit of 1+ 1/x as x goes to infinity and then say that you are taking $1^\infty$. The limits must be taken simultaneously.

How you show that $\lim_{x\to \infty}(1+ 1/x)^x= e$ depends upon exactly how you define e itself. In some texts, e is defined as that limit, after you have proved that the limit does, in fact, exist.

But you can also prove, without reference to e, that the derivative of the function $f(x)=a^x$ is a constant (depending on a) time $a^x$. And then define e to be such that that constant is 1.

That is, if $f(x)= a^x$ then $f(x+h)= a^{x+h}= a^xa^h$ so that
$$\frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h}$$
so that
$$\frac{da^x}{dx}= a^x \lim_{h\to 0}\frac{a^h- 1}{h}$$
and e is defined to be the number such that
$$\lim_{h\to 0}\frac{a^h- 1}{h}= 1$$.

That means that, for h sufficiently close to 0, we can write
$$\frac{a^h- 1}{h}$$
is approximately 1 so that
$$a^h- 1$$
is approximately equal to h and then $a^h$ is approximately equal to 1+ h.
That, in turn, means that a is approximately equal to $(1+ h)^{1/h}$

Now, let x= 1/h so that becomes $(1+ 1/x)^x$ and as h goes to 0, h goes to infinity.