Oscillation multiplied by a vanishing amplitude
= Oscillation multiplied by a vanishing amplitude
If $|h(x)|\leq C$ near $c$ and $a(x)\to0$ as $x\to c$, then $a(x)h(x)\to0$ by the <squeeze theorem>, even if $h$ has no limit. Defining the product to be zero at $c$ gives continuity there. For example, $\sqrt{|x|}\sin(1/\sin x)$ extends continuously at zero; the same oscillation near a nonzero multiple of $\pi$ has nonvanishing amplitude and does not have a continuous zero extension.