Problem with the definition of $e$?

Question

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?

Answer 1

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 $x \to 0$.

Answer 2

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 $A\to e^1=e$.