# How to prove that $(1+x)^r$ behaves like $1+rx$ for small x without calculus?

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.

• 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 Sep 21 '19 at 15:44
• 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 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

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.$$|x_n-r|<\delta\implies\left|f(x_n)-f(r)\right|=\lim_{y\to1}\left|\frac{y^{x_n}(y^{r-x_n}-1)}{y-1}\right|\le\delta.$$

• 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
• @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

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.