Compute
![\[ \lim_{x \to 0} x \sin \frac{1}{x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-47b5042b0474fcc4f360913d988080ec_l3.png "Rendered by QuickLaTeX.com")
Since 
for all 
we have
![\[ -x \leq x \sin \frac{1}{x} \leq x \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-09823eba71bb2ad892b0859516d2212b_l3.png "Rendered by QuickLaTeX.com")
for all . Since 
, we have
![\[ \lim_{x \to 0} x \sin \frac{1}{x} = 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ff6c48654cfa60b311b194d4fabeebf0_l3.png "Rendered by QuickLaTeX.com")
by the squeeze theorem (Theorem 3.3 in Apostol).
Compute
Since for all we have
for all . Since , we have
by the squeeze theorem (Theorem 3.3 in Apostol).
You have a mistake here. When you multiply your first inequality by x you need to consider the sign of x.
Right. That’s why it is better to use the inequalities
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