Oscillation multiplied by a vanishing amplitude

If $|h(x)|\le C$ near $c$ and $a(x)\to 0$ as $x\to c$, then $a(x)h(x)\to 0$ 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.