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

Question

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.

Answer 1

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

Answer 2

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.