# Napier's Constant Limit Definition

by scorsesse
Tags: constant, definition, limit, napier
 P: 1 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 ?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 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.

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