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I have an issue understanding one of the definitions of $e$ that I found in a textbook I am using. They defined e as the limit of $(1+x)^{1/x}$ as $x\to 0$.

But as $x$ approaches $0$ it can come in from either side of zero resulting in $1/x$ jumping back and forth between infinity and negative infinity. So I don't see how this is a valid definition.

Any help?

Larara
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B flat
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    This definition can also be found on wikipedia... so there must be a good reason why I am incorrect in my reasoning. – B flat Mar 04 '16 at 00:44
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    Indeed $1/x$ can be very large positive or very large negative. In either case, the expression will be close to $e$. (A more usual definition of $e$ is the limit as $n\to\infty$ of $(1+1/n)^n$.) – André Nicolas Mar 04 '16 at 00:50
  • @AndréNicolas: Indeed and in that limit $n$ is a positive natural number, which is better because it only requires natural number exponents. The definition with real $x \to 0$ is slightly problematic because it's harder to prove the existence of that limit without already knowing a lot about $e$. – user21820 Mar 04 '16 at 01:05

2 Answers2

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As $x \to 0$, $\frac{1}{x}$ does not jump between any kinds of infinities, because $x$ is never $0$. If however you're asking why it seems that it can work even when $\frac{1}{x}$ can be made to oscillate between arbitrarily large positive and negative values, then it is because that is merely one part of the expression for which you are taking the limit. The other part $(1+x)$ oscillates between more than and less than $1$ if $x$ is made to oscillate between positive and negative, and in this case $(1+x)^\frac{1}{x}$ tends to $e$ regardless of the oscillations of the individual parts.

$(1+x)^\frac{1}{x} = \exp(\frac{1}{x}\ln(1+x)) \in \exp(\frac{1}{x}(x+o(x))) \subseteq \exp(1+o(1)) \subseteq \exp(1)\exp(o(1))$

$\subseteq \exp(1)(1+o(1)) \subseteq \exp(1)+o(1)$ as $x \to 0$.

user21820
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    Don't forget $x \frac{1}{x} \to 1$ as $x \to 0$, despite the fact that $\frac{1}{x}$ can oscillate between arbitrarily large positive and negative values as $x \to 0$. – user21820 Mar 04 '16 at 00:51
  • user 21820, should the last part of your working be exp(1) ? Not sure if I understand the notation correctly. – Conan Wong Mar 04 '16 at 00:53
  • @ConanWong: You are right of course; my careless mistake. – user21820 Mar 04 '16 at 00:54
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Consider $$A=(1+x)^{1/x}$$ Take logarithms of both sides $$\log(A)=\frac 1x\log(1+x)$$ Now, consider Taylor series $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ which makes $$\log(A)=1-\frac{x}{2}+O\left(x^2\right)$$ Since $x\to 0$, $\log(A)\to 1$ and $A\to e^1=e$.

Claude Leibovici
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