need help with proof of simple theorem - linear algebra / analysis i'm stalled in my attempt to prove that $(1-a^2)^{\frac{1}{2}} - (1-b^2)^{\frac{1}{2}}$ goes to zero faster than $|A-B|$, where A,B are vectors in $\mathbb{R}^n$ with $a=|A| \leq 1$ and $b=|B|\leq 1$. Can anybody give me assistance? thanks if you can help
peace
stm
 A: Hint: consider a function $f(x)  =\sqrt{1-x^2}$ for real $x$ and calculating first two derivatives show that 
$$
|f(0) - f(x)| = O(x^2).
$$
If you need additional hints - please let me know.
A: Assuming $A \neq B$, use the reverse triangle inequality to get $|A - B| \geq |a-b|$, whence
$$\left|\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{|A-B|}\right| \leq \left|\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{a-b}\right|.$$
Thus you can conclude, for example,
$$\limsup_{B \to A} \left|\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{|A-B|}\right| \leq \frac{a}{\sqrt{1-a^2}}.\tag{1}$$
Now, consider how $B$ may approach $A$.  If $A$ and $B$ are parallel with, say, $b < a$ then $|A-B| = a-b$, so that
$$\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{|A-B|} = \frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{a-b} \to -\frac{a}{\sqrt{1-a^2}}$$
as $B \to A$ since the expression you get is just the difference quotient for the function $\sqrt{1-x^2}$.  In this case you can't say that the numerator goes to zero faster or slower than the denominator; in fact, they go to zero at the same rate in some sense (and that sense is known as big-$\Theta$).
And, if $a = b$ (that is, if $B$ approaches $A$ on the sphere of radius $a$) then the numerator is zero always.
By inequality $(1)$, every other path of approach must give you a behavior somewhere between these two.  The numerator will always go to zero at least as fast as $|A-B|$, and in some cases--but not all--it will go to zero faster.
