Showing that $\sqrt{1+x}=1+\tfrac12x+o(x)$ I have found that a nice way to explain to students what the derivative is the number $A$ such that
$$f(x+dx)=f(x) + A\,dx+o(dx),$$
motivated by the idea of approximating $f$ around $x$.
If we take this as the starting point of what differentiation is, it's easy to obtain things like $(2+dx)^2 \approx 4 + 4\,dx$ for small $x$, and so on for more general polynomials. But it would be nice to give a more non-trivial example. It feels like the next natural step is something like
$$\sqrt{1+dx} = 1 + \tfrac12dx+o(dx),$$
but how can I show this without using &quot;heavy machinery&quot;? At this point I've only started the topic, and my students have only seen $\epsilon$-$\delta$ style proofs of some basic limits (plus I mentioned some of the standard ones like $\sin x/x\to 1$ as $x\to 0$).
This is essentially an easy case of the more general power rule for derivatives: $d(x^r)=rx^{r-1}\,dx$, but all proofs I've seen use logarithms or the inverse function theorem (changing $f(x)=x^{1/2}$ to $f^{-1}(y)=y^2$ etc.). Is there a more direct way I can show that
$$\lim_{dx\to0}\frac{\sqrt{1+dx}-(1+\tfrac12\,dx)}{dx}=0?$$
Or maybe $\sqrt{1+x}$ is not the most natural &quot;next step&quot; up from polynomials, and there is another easier example I can give?
You may say that
$$
\sqrt {1 + dx}  - \left( {1 + \frac{1}{2}dx} \right) = \frac{{1 + dx - \left( {1 + \frac{1}{2}dx} \right)^2 }}{{\sqrt {1 + dx}  + \left( {1 + \frac{1}{2}dx} \right)}} = \frac{{ - \frac{1}{4}(dx)^2 }}{{\sqrt {1 + dx}  + \left( {1 + \frac{1}{2}dx} \right)}} = o(dx)
$$
as $dx \to 0$.
Using a derivative :
$$\lim_{h\to0}\frac{\sqrt{1+h}-1-\frac{h}{2}}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h-0}=f'(0)=0$$
where $f(h):=\sqrt{1+h}-\frac{h}{2}$.