TITLE: Suppose we want to make $|x^2 \sin(1/x)| < \varepsilon$ where $\varepsilon > 1$, why does it not suffice that $|x| < \varepsilon$?
QUESTION [5 upvotes]: I was reading Spivak's Calculus and came across this (Spivak's Calculus Chapter 5)
I'm confused because $f(x) = x^2 \sin(\frac{1}{x})$ has the following property: $|f(x)| < |x|$ for all $x ∈ \Bbb{R}$. If we set $|x| < \varepsilon$, then $|f(x)| < |x| < \varepsilon$ therefore $|f(x)| < \varepsilon$. Shouldn't it then suffice to set $|x| < \varepsilon$ if we want to make $|x^2 \sin(\frac{1}{x})| < \varepsilon$?

REPLY [5 votes]: Well, it does suffice to require $|x| < \varepsilon$, since (as you say) it is true that $|f(x)| \le |x|$ for all $x$. However, it is not completely obvious that this is true; you need to use more information about the factor $\sin(1/x)$ than just the trivial property $|\sin(1/x)| \le 1$. To avoid having to think about this, it's easier to just take $|x| < \min(\varepsilon,1)$ so that you can estimate as follows: $|f(x)| \le |x|^2 \le |x|$.