# Suppose we want to make $|x^2 \sin(1/x)| < \varepsilon$ where $\varepsilon > 1$, why does it not suffice that $|x| < \varepsilon$?

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 $|x^2 \sin(\frac{1}{x})| < \varepsilon$?

• it is not clear to me what is your problem. Isn't exactly what is written in your link (that it is is enough to require $|x|<\epsilon$)?
– Surb
Dec 16 '17 at 8:26
• Spivak stated the following: "If we are given an E > 1, then it does not suffice to require that |x| < E, but it certainly suffices to require that |x| < 1 and x =/= 0. I understand why it suffices to require that |x| < 1 and x =/= 0 but I do not understand why it does not suffice to require that |x| < E. Dec 16 '17 at 8:30

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: $|f(x)| \le |x|^2 \le |x|$.
• Just an example, let $\varepsilon=5$ and $|x|<\varepsilon$, then $f(x)<25$ but it's not very clear that $f(x)\leq4.96<\varepsilon=5$ Oct 21 '20 at 8:06