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Of course the fact that, in the neighborhood of $x=0$, $$ (1+x)^r=1+rx+o(x) $$ can be easily proven for integer $r$. For positive values, it's a trivial consequence of the binomial formula. For negative values it is possible to use the formula for geometric progression.

What about non integer $r$? I know everything about Taylor expansion, but I am looking for a non calculus proof. Maybe it is trivial, or there is something on this site, but I was not able to find it.

GiorgioP-DoomsdayClockIsAt-90
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    For the rational case you can use the integer result: suppose $(1+x)^{m/n}=1+ax+o(x)$ for some $a$, then $(1+x)^m=(1+ax)^n+o(x)$ but now $(1+ax)^n=1+anx+o(x)$, so $an=m$. This is basically like implicit differentiation. For the real case it is not apparent how $(1+x)^r$ is defined in the first place. – Ian Sep 21 '19 at 15:33
  • @Ian real case: let's assume that in some way it has been introduced the exponential function and its inverse (the logarithm), as usually done at pre-calculus level, so that $(1+x)^r = exp(r log(1+x))$. Is it possible to make progress from there? – GiorgioP-DoomsdayClockIsAt-90 Sep 21 '19 at 15:44
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    If you know something about expansion of the exponential function and the logarithm at $0$ and $1$ respectively then yes, but at that point you are getting pretty close to calculus. An approach that needs less calculus is to postulate continuity in $r$, in which case the result follows freely from the rational case. – Ian Sep 21 '19 at 15:45
  • @Ian Probably your suggestion about the extension to rational $r$ could be enough to build something which, if not a real proof, could be at least a good argument to justify the formula. Thanks. – GiorgioP-DoomsdayClockIsAt-90 Sep 21 '19 at 15:49
  • Well, it's a "proof" once you decide that $x^r$ should be continuous in $r$ for $x>0$ fixed, but then you're really just passing the buck off to justifying that assumption. (One milder assumption that implies this is that $x^r$ is monotone in $r$ for $x>0$ fixed. But again, you have to take something for granted to make sense of $x^r$ for real $r$.) – Ian Sep 21 '19 at 15:50
  • @Ian Well, it is a "proof" with quotation marks because I would have hard time to consider continuity a pre-calculus tool. However, it is true that even without the whole machinery, the concept of real numbers as objects which can be approximated better and better by sequences of rational numbers is often used even at K12 level. – GiorgioP-DoomsdayClockIsAt-90 Sep 21 '19 at 16:04
  • It may sound somewhat arbitrary that I consider calculus to include derivatives and integrals but not limits and continuity, which to me are analysis instead. But if this is granted, defining $y^r$ to be continuous in $r$ is straightforward. No $\exp$, no $\log$. Why do I define calculus so? Because calculus was historically difficult due to us trying to think about slopes of curves before we'd really dealt with limits adequately, so I take calculus to start where it did historically, excluding what should have been done first but wasn't. – J.G. Sep 21 '19 at 16:25

2 Answers2

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We need a formal statement of what we want to prove, namely $f(r):=\lim_{y\to1}\frac{y^r-1}{y-1}=r$. You already noted the binomial theorem covers positive (in fact, non-negative) integers $r$, and @Ian notes we get the rational case easily, viz. $$m,\,n\in\Bbb Z,\,n>0\implies f(m/n)=\frac{\lim_{x\to0}\frac{y^m-1}{y-1}}{\lim_{x\to0}\frac{y^m-1}{y^{m/n}-1}}=\frac{f(m)}{f(n)}=\frac{m}{n}.$$We then cover irrational $r$ by checking $f$ is continuous. Indeed, if a sequence $x_n$ satisfies $\lim_{n\to\infty}x_n=r$, and we define $y^r$ to be $r$-continuous so $f$ is increasing,$$|x_n-r|<\delta\implies\left|f(x_n)-f(r)\right|=\lim_{y\to1}\left|\frac{y^{r}(y^{x_n-r}-1)}{y-1}\right|=|f(x_n-r)|<\Delta$$for all rationals $\Delta>\delta$, i.e.$$|x_n-r|<\delta\implies\left|f(x_n)-f(r)\right|=\lim_{y\to1}\left|\frac{y^{x_n}(y^{r-x_n}-1)}{y-1}\right|\le\delta.$$

J.G.
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  • Is continuity of $f$ clear using a non-calculus definition of $y^r$? It seems the claim involves an interchange of limits. – jawheele Sep 21 '19 at 16:52
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    @jawheele As in $\lim_{y\to1}\frac{y^{\lim_{n\to\infty}q_n}-1}{y-1}=\lim_{y\to1}\lim_{n\to\infty}\frac{y^{q_n}-1}{y-1}\stackrel{interchange}{=}\lim_{n\to\infty}\lim_{y\to1}\frac{y^{q_n}-1}{y-1}$ for any sequence $q_n$ of rationals with $\lim_{n\to\infty}q_n=r$? Well-spotted. I might address that in an edit later. – J.G. Sep 21 '19 at 16:55
  • Indeed, and to extend the result by continuity to $\mathbb{R}$, you'd need to show continuity on $\mathbb{R}$, i.e. the interchange should work for any sequence of reals converging to $r$. – jawheele Sep 21 '19 at 17:07
  • @jawheele I've given it a try. – J.G. Sep 21 '19 at 17:14
  • Nice! Though shouldn't $|f(x_n)-r|$ be replaced with $|f(x_n)-f(r)|$ to show continuity? – jawheele Sep 21 '19 at 18:00
  • @jawheele A typo! Thanks; fixed. – J.G. Sep 21 '19 at 18:01
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In the non-rational case, one has to confront the relevant definition of exponentiation: $$a^b := e^{b\ln{a}}, \;\; a>0, b \in \mathbb{R}$$ Oftentimes the construction of the exponential function and natural logarithm is based in calculus, but one can avoid it by defining $e^x$ by its power series and $\ln(x)$ as the inverse. Using this approach, we proceed as follows:

\begin{align}(1+x)^r-(1+rx) & = e^{r\ln(1+x)} -(1+rx) \\ &= \sum_{k=0}^\infty \frac{r^k\ln(1+x)^k}{k!} - (1+rx) \\ & = r\ln(1+x)-rx + \sum_{k=2}^\infty \frac{r^k \ln(1+x)^k}{k!}. \end{align}

One would like to use l'Hopital's rule to show that $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$, from which it would follow that the RHS is $o(x)$, our desired result. If we wish to avoid calculus entirely, we must justify this limit otherwise. Exponentiating and using our definition, however, this is equivalent to showing the well-known result

$$\lim_{x \to 0} (1+x)^{1/x} = e := \sum_{k=0}^\infty \frac{1}{k!}.$$

If you believe this limit, you're done, but it sounds rather bothersome to prove it without a calculus-based definition of the exponential function or natural logarithm while avoiding circular reasoning. What one can certainly do without calculus is prove that the equality holds for the sequence $x_n=\frac{1}{n}$ (see this answer), but this is not sufficient for the desired result. You could probably modify the proof for that sequence to work for any sequence of rational numbers approaching $0$; I'd imagine this is sufficient for our limit to hold by the density of the rational numbers, a la this result.

jawheele
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