How do you prove lim h->0 of (1+h)^(1/h) exists?

I am planning to teach calculus to a student, and I would like to define Euler's number, $e$ , as $\lim \limits_{h\to0} (1+h)^{\frac1h} .$ However, I do not know how to show that this limit exists (and is finite). Any help is appreciated!

#Calculus

Easy Math Editor

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Comments

Start by proving that $x - \tfrac12x^2 < \ln(1+x) < x$ for all $x > 0$ .

If $f(x) = x - \ln(1+x)$ , then $f'(x) = \tfrac{x}{1+x} > 0$ for all $x > 0$ and $f(0) =0$ . The MVT gives $f(x) >0$ for $x > 0$ .
If $g(x) = \ln(1+x)-x+\tfrac12x^2$ , then $g'(x) = \tfrac{x^2}{1+x} > 0$ for all $x > 0$ , and $g(0)= 0$ . The MVT gives $g(x) > 0$ for $x > 0$ .

Then $-\tfrac12x < x^{-1}\ln(1+x) - 1 < 0$ for $x > 0$ , and hence $\lim_{x \to 0+}x^{-1}\ln(1+x) = 1$ . Given that exponentiation is continuous, we get the result.

However, we need to know about exponentiation and logarithms, and this comes a little after the existence of $e$ , unless you take the approach below...

Define $e^x$ by its power series $E(x) \; = \; \sum_{n=0}^\infty \frac{1}{n!}x^n$ and then show that:

this power series has infinite radius of convergence,
as such a power series, $E$ is differentiable and equal to its own derivative,
as a consequence of this, use the MVT to show that $E(a+b)=E(a)E(b)$ for $a,b \in \mathbb{R}$ ,
show that $E$ is a strictly increasing bijection from $\mathbb{R}$ to $(0,\infty)$ , and deduce the existence of its inverse function $L\,:\, (0,\infty) \to \mathbb{R}$ , which has the properties that $L(xy) = L(x)+L(y)$ for $x,y > 0$ and $L'(x) = x^{-1}$ .
define $e = E(1)$ , and you now have all the tools needed to follow the original argument.

Your student would need a good foundation in basic analysis (MVT, Intermediate Value Theorem, Inverse Function Theorem, differentiability of power series) to follow this route.

Mark Hennings - 4 months, 4 weeks ago

I just remembered I found another method online to use $\lim_{h\to 0} (1+h)^\frac{1}{h} = e$ to prove $\lim_{h\to 0} \frac{e^h-1}{h}=1,$ but any new answers are welcome.