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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!!

kaithkolesidou
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    You can't apply de l'Hopital: the numerator does not approach to $\infty$. What you can do is to prove that $(\sin x - s \cos x)$ is bounded, and that $$\lim_{x \to +\infty} xe^{-sx}=0$$ – Crostul Apr 26 '16 at 13:02
  • I just did it.. it was such a stupid mistake! Thank you very much! – kaithkolesidou Apr 26 '16 at 13:08

2 Answers2

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First, is it true that $s>0$ and $s$ is fixed?

If so, you can prove first using l'Hopital that $\displaystyle{\lim_{x\to\infty} x\cdot e^{-sx}}=0$, and then use the fact that $|x\cdot e^{-sx}(\sin x-s\cos x)|\leq (s+1)|x\cdot e^{-sx}|$ to show that

\begin{align*} 0&\leq \left|\lim_{x\to\infty} |x\cdot e^{-sx}(\sin x-s\cos x)|\right|\\ &=\lim_{x\to\infty} |x\cdot e^{-sx}(\sin x-s\cos x)|\\ &\leq (s+1)\cdot \lim_{x\to\infty} x\cdot e^{-sx}=0 \end{align*}

This implies in particular that $\displaystyle{\lim_{x\to\infty} x\cdot e^{-sx}(\sin x-s\cos x)}=0.$

Darío G
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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 $0$

Archis Welankar
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