Approximate the function $\sqrt{1-x^2}$ for $x\rightarrow0$ in a "physics context" So, to reiterate, we need to approximate the function $f(x)=\sqrt{1-x^2}$ for $x\rightarrow0$.
This comes from a physics problem (trying to make an approximation in the phonon dispersion relation) and what it really wants, is a Taylor expansion and keep up to 1st order terms.
The problem lies in the following: if your take the Taylor expansion at $x=0$, the linear term goes away and you lose all the physics. My question is if there is a workaround this issue. For example, i thought if we first approximate $x^2$ with $x$ and then do the Taylor expansion, the problem goes away, but that seems so ad-hoc. Any suggestion would be welcome.
 A: The fact that the linear term dissappears is exactly what we want. Let's take a brief look at the plot of this approximation

We see that around $x=0$ the function is flat (the derivative is zero). This observation is encapsulated in the linear term being zero. So this is the physics you want. If the linear term in a Taylor expansion is zero this means you are either in a maximum or a minimum and this is used many times in physics (although sometimes in more sophisticated forms).
In general it can happen that one or more terms are zero in a taylor expansion. Especially when the function under consideration is symmetric. In this case the function is even so $f(x)=f(-x)$. You can proof for yourself that every odd power in the expansion must be zero by imposing $\sum_n a_n x^n=\sum_n a_n (-x)^n$
A: $$
f(x)=\sqrt{1-x^2}\approx 1-\frac{x^2}{2}
$$
for small $x$.  There is no linear term and you can’t “make one appear”.  Your function does not have a linear part near $x=0$.  If you want to have a linear term you need to consider something like
$$
f(x)=\sqrt{1-(x-x_0)^2}\approx 1-\frac{1}{2}(x_0^2-2x_0x)
$$
but that’s not expanding near $x=0$.
