# Math Help - Question regarding limits. Help appreciated!

1. ## Question regarding limits. Help appreciated!

what is lim(x,y) approaching zero sinxcos(1/y) by using squeeze theorem?

2. ## Re: Question regarding limits. Help appreciated!

$-1 \leq cos(\frac{1}{y}) \leq 1$ and so
tiny condondrum to the above as sin(x) can take on negative values.

So more apt would be either
$-sin(x) \leq sin(x) cos(\frac{1}{y}) \leq sin(x)$
or
$-sin(x) \geq sin(x) cos(\frac{1}{y}) \geq sin(x)$

either way the limit is the same for sin(x) = 0 $x \to 0$