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Exponential valueby henrywang
Tags: exponential 
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#1
Jul714, 02:49 PM

P: 13

For proving the natural number, e.
(1+1/n)^n As n approches infinite, (1+1/n)^n > e However, wouldnt it become one as n becomes infinite? (1+1/n)^n=(1+0)^n=1 Could anyone explain this to me please! 


#2
Jul714, 02:56 PM

Mentor
P: 18,036

If you just naively substitute ##n=\infty##, then you would get
[tex](1+0)^\infty = 1^\infty[/tex] which is an indeterminate form. Surely, things like ##1^2## and ##1^{100}## equal ##1##. But if the exponent is infinity, then it's an indeterminate form. 


#3
Jul714, 03:37 PM

Sci Advisor
P: 6,037

[tex](1+\frac{1}{n})^n = 1 + n\frac{1}{n} +\frac{ n(n1)}{2!} \frac{1}{n^2} + ... \frac{1}{n^n}[/tex]
This does not >1 as n becomes infinite. 


#4
Jul814, 07:12 AM

P: 907

Exponential value
[itex] \displaystyle\lim_{n\rightarrow +\infty} {(1+\frac{1}{n})^n} [/itex] The attempted manipulation was to: [itex] \displaystyle\lim_{{n_1}\rightarrow +\infty} {\displaystyle\lim_{{n_2}\rightarrow +\infty} {(1+\frac{1}{n_2})^{n_1}}} [/itex] which is distinct from: [itex] \displaystyle\lim_{{n_2}\rightarrow +\infty} {\displaystyle\lim_{{n_1}\rightarrow +\infty} {(1+\frac{1}{n_2})^{n_1}}} [/itex] Those two nested limits give different results. The first one evaluates to 1. The second one does not exist at all. Limits do not always commute. One way of seeing this is to imagine the formula [itex]{(1+\frac{1}{x})^y}[/itex] as a function of two variables. It is defined everywhere in the first quadrant of the coordinate plane for x and y. You can imagine extending the definition to infinite x and infinite y by tracing a path going off into the distance and looking at the limit of the function values along that path. It then turns out that the limit you get depends on the path you take. The prescription in the original formulation was for a path running exactly on the 45° line through the middle of the first quadrant. 


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