<s>[INST] Evaluating the magnitude. Is this less than 1? Where $|x|<1$, I'm looking to determine if
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right|<1$$
I believe it is, since we can use a Taylor series to approximate
$$\sqrt{1-x^2} = 1 - \frac{1}{2}x^2-\frac{1}{4}(x^2)^2 + O(3) \approx 1 - \frac{1}{2}x^2$$
This then gives
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right| \approx \left|\frac{1}{x}(-1+1 - \frac{1}{2}x^2)\right| = \left|-\frac{x}{2}\right|<1$$
This, though, is not a proof. Any ideas?
 [/INST] HINT:
Let $2y=\arcsin x,0\le2y\le\dfrac\pi2$
$\implies x=\sin2y,\cos2y=+\sqrt{1-x^2}$
$$\dfrac{\sqrt{1-x^2}-1}x=\dfrac{\cos2y-1}{\sin2y}=\cdots=-\tan y$$
Now $$0\le y\le\dfrac\pi4\implies0\le\tan y\le1$$
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