Napier's Constant Limit Definition

[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 ?

 

[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.