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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.

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    $\begingroup$ 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. $\endgroup$ – Ian Sep 21 at 15:33
  • $\begingroup$ @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? $\endgroup$ – GiorgioP Sep 21 at 15:44
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    $\begingroup$ 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. $\endgroup$ – Ian Sep 21 at 15:45
  • $\begingroup$ @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. $\endgroup$ – GiorgioP Sep 21 at 15:49
  • $\begingroup$ 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 $r$.) $\endgroup$ – Ian Sep 21 at 15:50
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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.$$|x_n-r|<\delta\implies\left|f(x_n)-f(r)\right|=\lim_{y\to1}\left|\frac{y^{x_n}(y^{r-x_n}-1)}{y-1}\right|\le\delta.$$

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  • $\begingroup$ Is continuity of $f$ clear using a non-calculus definition of $y^r$? It seems the claim involves an interchange of limits. $\endgroup$ – jawheele Sep 21 at 16:52
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    $\begingroup$ @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. $\endgroup$ – J.G. Sep 21 at 16:55
  • $\begingroup$ 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 $r$. $\endgroup$ – jawheele Sep 21 at 17:07
  • $\begingroup$ @jawheele I've given it a try. $\endgroup$ – J.G. Sep 21 at 17:14
  • $\begingroup$ Nice! Though shouldn't $|f(x_n)-r|$ be replaced with $|f(x_n)-f(r)|$ to show continuity? $\endgroup$ – jawheele Sep 21 at 18:00
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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.

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