Simplification of $ \sqrt{(1-x^2)}$ to $(1-\frac{x^2}{2})$ While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:
$$
\sqrt{1-x^2} \approx \left(1-\frac{x^2}{2}\right)
$$
I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).
 A: Term to look for: linear approximation 
In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) \approx f(c) + f'(c)\;(x-c)
$$
This is essentially the definition of the derivative.  And you should find this in your calculus book soon after the definition of derivative.  
Now if $f(x) = \sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-\frac{1}{2}$.  So
$$
\sqrt{1-x} \approx 1 - \frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.
A: Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).
$$\sqrt{1-t}\approx 1-\frac t2.$$
Hence for small $x$,
$$\sqrt{1-x^2}\approx 1-\frac{x^2}2.$$

The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)
$$\sqrt{1-t}\approx 1-\frac t2-\frac{t^2}8,$$
and
$$\sqrt{1-x^2}\approx 1-\frac{x^2}2-\frac{x^4}8,$$

A: Use the Taylor series expansion  at $x=0$ to get $\sqrt{1-x^2}\approx1-\frac{x^2}2+o(x^4)$.
