Given that e is defined to be limn→∞(1+1n)n, how do I prove that e=limn→0(1+n)1n?

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There are a couple of good ways to go about this, but let’s try to prove the following general statement:

Given some function f$f$, the limit of f(x)$f(x)$ as x→∞$x\to \mathrm{\infty }$ is equal to the limit of f(1x)$f\left(\frac{1}{x}\right)$ as x→0+$x\to 0^{+}$.

Okay, so we start off with the assumption that limx→∞f(x)=L$\underset{x\to \mathrm{\infty }}{lim}f(x)=L$ for some L$L$. This means that given any ϵ>0$ϵ>0$, there exists some number Mϵ$M_{ϵ}$ such that |f(x)−L|<ϵ$|f(x)-L|<ϵ$ for all x>Mϵ$x>M_{ϵ}$.

We wish to show that, given any ϵ>0$ϵ>0$, there exists some number δϵ${\delta }_{ϵ}$ such that ∣∣f(1x)−L∣∣<ϵ$|f\left(\frac{1}{x}\right)-L|<ϵ$ for all 0<x<δϵ$0<x<{\delta }_{ϵ}$.

This should be pretty easy - simply choose δϵ=1Mϵ${\delta }_{ϵ}=\frac{1}{M_{ϵ}}$. If 0<x<1Mϵ$0<x<\frac{1}{M_{ϵ}}$ then 1x>Mϵ$\frac{1}{x}>M_{ϵ}$, and so ∣∣f(1x)−L∣∣<ϵ$|f\left(\frac{1}{x}\right)-L|<ϵ$ by our initial assumption. Because we can do this for all ϵ>0$ϵ>0$, we have that limx→∞f(x)=L⟹limx→0+f(1x)=L$\underset{x\to \mathrm{\infty }}{lim}f(x)=L⟹\underset{x\to 0^{+}}{lim}f\left(\frac{1}{x}\right)=L$.

From there, simply let f(x)=(1+1x)x$f(x)={(1+\frac{1}{x})}^x$ for your specific example, and the desired result follows.

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Note that the 0+$0^{+}$ is crucial here. If we had tried to prove this for the general limit as x→0$x\to 0$, then we would have had to seek some δϵ${\delta }_{ϵ}$ which would work for all 0<|x|<ϵ$0<|x|<ϵ$. The problem with this is that 0<|x|<δ$0<|x|<\delta$ does *not* imply that 1x>1δ$\frac{1}{x}>\frac{1}{\delta }$ - it implies that either 1x>1δ$\frac{1}{x}>\frac{1}{\delta }$ or 1x<−1δ$\frac{1}{x}<-\frac{1}{\delta }$. Because our initial assumption is related only to the former case, then in general the statement is not true (unless of course we also have that limx→−∞f(x)=limx→∞f(x)=L$\underset{x\to -\mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}f(x)=L$).