what is lim(x,y) approaching zero sinxcos(1/y) by using squeeze theorem?
$\displaystyle -1 \leq cos(\frac{1}{y}) \leq 1 $ and so
tiny condondrum to the above as sin(x) can take on negative values.
So more apt would be either
$\displaystyle -sin(x) \leq sin(x) cos(\frac{1}{y}) \leq sin(x) $
or
$\displaystyle -sin(x) \geq sin(x) cos(\frac{1}{y}) \geq sin(x) $
either way the limit is the same for sin(x) = 0 $\displaystyle x \to 0 $