Showing that $\sqrt{1+x}=1+\tfrac12x+o(x)$

Question

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 "heavy machinery"? 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 "next step" up from polynomials, and there is another easier example I can give?

Answer 1

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 $dx \to 0$.

Answer 2

Do they know Newton's approximations, which are a consequence of the binomial series? If yes, you can easily show using that series. Since $dx$ is small, you neglect the higher powers, and the desired expression is obtained. The limit you wish to evaluate can also be evaluated using said method.

Answer 3

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 $f(h):=\sqrt{1+h}-\frac{h}{2}$.

Answer 4

It can be useful to obtain the derivative "from scratch" as follows: we look for the value of $\alpha$ that minimizes the error

$$\sqrt{x+1}-1-\alpha x$$ for small $x$.

We can rework the expression as follows:

$$\sqrt{x+1}-1-\alpha x=\frac{x+1-(1+\alpha x)^2}{\sqrt{x+1}+1+\alpha x}=\frac{x-2\alpha x+\alpha^2x^2}{\sqrt{x+1}+1+\alpha x}\sim x\frac{1-2\alpha+\alpha^2x}{2}.$$ The denominator $2$ is justified by the fact that for smaller and smaller $x$, $x$ becomes negligible. In the numerator, $\alpha^2x$ could also be considered negligible, unless the constant term vanishes. Indeed, with $\alpha=\dfrac12$, the error is virtually

$$\frac{x^2}8=o(x).$$