
#1
Mar3112, 08:41 AM

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 ? 



#2
Mar3112, 09:29 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

You cannot first take the limit of 1+ 1/x as x goes to infinity and then say that you are taking [itex]1^\infty[/itex]. The limits must be taken simultaneously.
How you show that [itex]\lim_{x\to \infty}(1+ 1/x)^x= e[/itex] 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 [itex]f(x)=a^x[/itex] is a constant (depending on a) time [itex]a^x[/itex]. And then define e to be such that that constant is 1. That is, if [itex]f(x)= a^x[/itex] then [itex]f(x+h)= a^{x+h}= a^xa^h[/itex] so that [tex]\frac{a^{x+h} a^x}{h}= \frac{a^xa^h a^x}{h}= a^x\frac{a^h 1}{h}[/tex] so that [tex]\frac{da^x}{dx}= a^x \lim_{h\to 0}\frac{a^h 1}{h}[/tex] and e is defined to be the number such that [tex]\lim_{h\to 0}\frac{a^h 1}{h}= 1[/tex]. That means that, for h sufficiently close to 0, we can write [tex]\frac{a^h 1}{h}[/tex] is approximately 1 so that [tex]a^h 1[/tex] is approximately equal to h and then [itex]a^h[/itex] is approximately equal to 1+ h. That, in turn, means that a is approximately equal to [itex](1+ h)^{1/h}[/itex] Now, let x= 1/h so that becomes [itex](1+ 1/x)^x[/itex] and as h goes to 0, h goes to infinity. 


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