Misunderstanding about the definition of a limit (Spivak Calculus)

Question

In Spivak's text, I quote:

"In general, if $\epsilon > 0$ to ensure that $|x^2\sin(\frac{1}{x})| < \epsilon$ we need only require that $|x| < \epsilon$ and $x \ne 0$"

This can easily be proven false if $|x| < |x^2\sin(\frac{1}{x})|$ Then how is it true; let's try to work it out.

$|x^2\sin(\frac{1}{x})| < \epsilon$ means $-\epsilon < x^2\sin(\frac{1}{x}) < \epsilon$

$|x| < \epsilon$ means $-\epsilon < x < \epsilon$

How does this then prove the latter idea?

Answer 1

It says "to ensure that $|x^2\sin(1/x)|<\varepsilon$ we need ONLY require that $|x|<\varepsilon$ and $x\ne0$" (emphasis added).

What that actually means is that IF $|x|<\varepsilon$ and $x\ne0$ THEN $|x^2\sin(1/x)|<\varepsilon$.

Notice that $|x^2\sin(\cdots)|\le|x^2|<|x|$ if $|x|<1$ (since $|\sin(\cdots)|\le1$). So if $|x|<\varepsilon$ then $|x^2\sin(1/x)|<\varepsilon$ unless $x=0$.

Answer 2

The missing part is

$$\left\lvert x \sin \left(\tfrac{1}{x}\right)\right\rvert \leqslant 1$$

for $x \neq 0$. That follows from the inequality $\lvert \sin y\rvert \leqslant \lvert y\rvert$ by setting $y = \frac{1}{x}$. Hence

$$\left\lvert x^2 \sin \left(\tfrac{1}{x}\right)\right\rvert \leqslant \lvert x\rvert$$

for $x\neq 0$.