I came across this problem in "Introduction to Real Analysis" by Bartle and had a doubt regarding the solution to a given problem. I needed to determine a condition on $|x-1|$ that will ensure that $$|(x^2-1)| < 1/2 $$
I started out by factorising : $|x^2-1| = |(x-1)(x+1)| <1/2 $
We have control over $|x-1|$ however we need to get rid of the $|x+1|$ factor to get a relation very similar to $|x-1| <\delta$.
This is where I am having a problem. In the examples given, they have generally assumed that $|x-1|<1$. By presupposing this condition, I am able to solve it but am not able to understand why 1 is being chosen since it seems like an arbitrary choice and the method should also work for some other value of $\delta$. This does not seem to be the case as if you take any other number, you do not get the correct condition. Could someone please explain the rational behind using only 1? Thank You.
People usually choose $1$ because that's the simplest choice. But any number greater than $0$ will do. I will do it with $4$. So, let us start by assuming that $\lvert x-1\rvert<4$. Then\begin{align}\lvert x+1\rvert&=\bigl\lvert(x-1)+2\bigr\rvert\\&\leqslant\lvert x-1\rvert+2\\&<6.\end{align}So, for any $\varepsilon>0$, take $\delta=\min\left\{4,\frac\varepsilon6\right\}$, and then\begin{align}\lvert x-1\rvert<\delta\implies\lvert x^2-1\rvert&=\lvert x-1\rvert\times\lvert x+1\rvert\\&<\frac\varepsilon6\times6\\&=\varepsilon.\end{align}