limit to infinity : trouble with l'hopital Given the following  limit for s positive constant
$\lim_{x\to \infty} xe^{-sx}(\sin x-s\cos x) $
how can I prove that the above is equal to $0$ ? 
I re-write the limit as  $ \frac{x(\sin x-s\cos x)}{e^{sx}} $  and then I use de l'Hopital theorem but it seems that I only go round and round.. 
I would appreciate any help! Thanks in advance!!

See $0\leq sin(x),cos(x)\leq 1$. So numerator is just oscillating between $x,0$ so now if you separate out you will see that denominator is growing so rapidly that at large x the value is almost $0$. So the limit as $x->\infty$ is $0$
