<s>[INST] Necessary condition on $|x-1|$ to assure $|x^2 -1| < \frac{1}{2}$. Is my solution correct? How to get the tightest answer? Question: Determine a condition on $|x-1|$ that will assure that:-
$$|x^2 -1| < \frac{1}{2}$$
My solution:-
Let $f(x) = x^2$
$$\lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} x^2 = 1$$
To find $\delta$ so that
$\: \: |f(x) - 1| < \epsilon = \frac{1}{2}\: \:$ whenever $\: \: 0 < |x-1| < \delta$
$$|f(x) - 1| < \epsilon = |x^2-1| = |x+1||x-1|$$
$$|f(x)-1| = |x+1||x-1|$$
$$|f(x)-1| < \delta \: |x+1| \tag {1}$$
$$$$
$$0 < |x-1| < \delta$$
$$|x-1| < \delta$$
$$-\delta < x-1 < \delta$$
$$-\delta +2 < x+1 < \delta + 2$$
$$-\delta-2 < x+1 < \delta + 2$$
$$|x+1| < \delta + 2 \tag{2}$$
Subsitutuing equation (2) in (1)
$$|f(x) -1| < \delta \: |x+1|$$
$$| f(x) - 1| < \delta \: (\delta + 2)$$
Solving $\delta \: (\delta + 2) = \frac{1}{2}$ for $\delta$ we get:
$$\delta = \frac{-2\pm \sqrt{6}}{2}$$
Since $\delta$ can only be positive, the necessary condition is
$$0 < |x-1| < \frac{-2+\sqrt{6}}{2} \approx 0.2247$$
 [/INST] $$|x^2 - 1|<\frac{1}{2}$$
$$-\frac{1}{2} < x^2 - 1<\frac{1}{2}$$
$$1 -\frac{1}{2} < x^2 < 1+ \frac{1}{2}$$
$$\sqrt{ \frac{1}{2} }< x < \sqrt{\frac{3}{2}} \text{ or } -\sqrt{ \frac{3}{2} }< x <-\sqrt{\frac{1}{2}}
$$
$$\sqrt{ \frac{1}{2} }-1< x-1 < \sqrt{\frac{3}{2}}-1 \text{ or } -\sqrt{ \frac{3}{2} }-1< x-1 <-\sqrt{\frac{1}{2}}-1
$$
A necessary (but not sufficient) condition is $$|x-1| < \sqrt{\frac{3}{2}}+1.$$
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