l'Hopital's rule needs to be applied. Any help will be greatly appreciated!
$\displaystyle \lim_{x \to \infty}{(x-sinx)(e^{-x^2})}$
1. Convert it to the form infinity/infinity and simplify
$\displaystyle \lim_{x \to \infty} \frac {x-sinx}{e^{x^2}}$
2. Take the derivatives
$\displaystyle \lim_{x \to \infty} \frac {1-cosx}{(e^{x^2})(2x)}$
3. Since 1 - cosx is either some constant or zero and the denominator is infinity, there are two cases. Either constant/infinity or zero/infinity both of which equal to zero.