Limits of Functions .... L&S Example 10.7 (2) ....

[Math Amateur]

I am reading "Real Analysis: Foundations and Functions of One Variable"by Miklos Laczkovich and Vera Sos ...

I need help with an aspect of Example 10.7 (2) ... Example 10.7 (2) reads as follows:View attachment 7252
In the above text, we read the following: "... ... Since whenever $$\lvert x - 2 \lvert \lt \frac{1}{2} , \lvert x - 1 \lvert \gt \frac{1}{2}$$ ... ... "Can someone please explain why:$$\lvert x - 2 \lvert \lt \frac{1}{2} \Longrightarrow \lvert x - 1 \lvert \gt \frac{1}{2}$$ ...Help will be appreciated ... Peter

 

[Opalg]

Can someone please explain why:$$\lvert x - 2 \lvert \lt \frac{1}{2} \Longrightarrow \lvert x - 1 \lvert \gt \frac{1}{2}$$ ...

More trickery with the triangle inequality! $$1 = |2-1| = |(2-x) + (x-1)| \leqslant |2-x| + |x-1| < \tfrac12 + |x-1|$$ and therefore $|x-1| > 1 - \frac12 = \frac12.$

Or to put it in everyday language, "the closer is to 2, the further is from 1".

 

[Math Amateur]

More trickery with the triangle inequality! $$1 = |2-1| = |(2-x) + (x-1)| \leqslant |2-x| + |x-1| < \tfrac12 + |x-1|$$ and therefore $|x-1| > 1 - \frac12 = \frac12.$

Or to put it in everyday language, "the closer is to 2, the further is from 1".

OMG ... I suppose I'll get the knack of these inequalities after a bit of practice :( ...

Thanks for your help ... you certainly took the mystery out of why the inequality held true ...

Peter

 

[HallsofIvy]

Equivalently, $$|x- 2|< 1/2$$ means that [math]-1/2< x- 2< 1/2[/math]. Adding 1 to each part, 1/2< x- 1< 3/2. Since x- 1 is always positive, we have |x- 1|= x- 1> 1/2.