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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$$

Arpit Rathore
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2 Answers2

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I like how you are trying to use the concept of continuity here. But the problem is much simpler. One can determine the necessary and sufficient condition on $|x-1|$ by simply solving the inequality directly.

First, an algebraic solution:

$$|x^2-1|<\frac{1}{2}\iff -\frac{1}{2}<x^2-1<\frac{1}{2} \iff \frac{1}{2}<x^2<\frac{3}{2}\iff \frac{1}{\sqrt{2}}<|x|<\frac{\sqrt{3}}{\sqrt{2}} $$ And $$\frac{1}{\sqrt{2}}<|x|<\frac{\sqrt{3}}{\sqrt{2}}\iff x\in ]-\frac{\sqrt{3}}{\sqrt{2}},\frac{\sqrt{3}}{\sqrt{2}}[\, \cap \left( ]\frac{1}{\sqrt{2}},\infty[\,\cap\, ]-\infty,-\frac{1}{\sqrt{2}}[ \right) $$ $$\iff x\in ]-\frac{\sqrt{3}}{\sqrt{2}},-\frac{1}{\sqrt{2}}[\,\cup\, ]\frac{1}{\sqrt{2}},-\frac{\sqrt{3}}{\sqrt{2}}[.$$

Second, a geometric solution:

You know that $x\mapsto x^2-1$ is a parabola whose axis of symmetry is the $y-axis$, vertex at $(-1,0)$, and open upwards. Find the points of intersection of the parabola with the two horizontal lines $y=-1/2$ and $y=1/2$. You immediately discover the two intervals where the curve $x^2-1$ lies over the line $y=-1/2$ and below the line $y=-1/2$.

Your original argument is not wrong. It is actually creative.

Medo
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  • Can you plz explain to me the bracket placements? Other than that I am able to understand your solution perfectly. – Arpit Rathore Aug 02 '21 at 03:36
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    $(a,b)$ is sometimes written as $]a,b[ $. – mjw Aug 02 '21 at 03:52
  • @mjw Thank you. – Arpit Rathore Aug 02 '21 at 04:07
  • From the working of @ArpitRathore in question, it appears the problem is related to limits. And solving inequalities is so so against the whole idea of limits. Compare these two questions: find a value of $x$ for which $x^2-5x+6=0$, and find all the values of $x$ for which $x^2-5x+6=0$. The definition of limit is not about solving inequalities but rather ensuring some inequalities and resemble finding a solution rather than all solutions. – Paramanand Singh Aug 03 '21 at 16:03
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$$|x^2 - 1|<\frac{1}{2}$$

$$-\frac{1}{2} < x^2 - 1<\frac{1}{2}$$

$$1 -\frac{1}{2} < x^2 < 1+ \frac{1}{2}$$

$$\sqrt{ \frac{1}{2} }< x < \sqrt{\frac{3}{2}} \text{ or } -\sqrt{ \frac{3}{2} }< x <-\sqrt{\frac{1}{2}} $$

$$\sqrt{ \frac{1}{2} }-1< x-1 < \sqrt{\frac{3}{2}}-1 \text{ or } -\sqrt{ \frac{3}{2} }-1< x-1 <-\sqrt{\frac{1}{2}}-1 $$

A necessary (but not sufficient) condition is $$|x-1| < \sqrt{\frac{3}{2}}+1.$$

mjw
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