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Let $$u(x_1,x_2)=\frac{1}{n^2}\sin(nx_1)\sinh(nx_2)$$

with $(x_1,x_2) \in \mathbb{R}^2$. What happens to $u$ as $n \to +\infty.$ ?

This is what I tried to do : $$-\frac{1}{n^2}\sinh(nx_2)\le u(x_1,x_2)\le\frac{1}{n^2}\sinh(nx_2)$$ but I don't know how to conclude.

Thanks.

mostafa
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  • Hello @mostafa, welcome to MSE. Can you add your attempts and efforts in the question (not the comments). Questions that don't show any efforts are sometimes poorly received on MSE. – Ernie060 Nov 11 '18 at 20:07

2 Answers2

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For $x_2=0 \, \lor \, x_1=k\pi$

$$\frac{1}{n^2}\sin(nx_1)\sinh(nx_2)=0$$

otherwise observe that

$$\sinh(nx_2) \sim \frac12e^{n|x_2|}\quad\lor\quad \sinh(nx_2) \sim -\frac12e^{n|x_2|}$$

but for $x_1\neq k\pi$ we have that $\sin (nx_1)$ doesn't converge.

Refer to the related

user
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  • I don't see how you prove that in the second case we don't have a limit ? can you please explain more? Thanks – mostafa Nov 11 '18 at 20:16
  • @mostafa Recall that $\sinh x=\frac{e^x-e^{-x}}{2}$ then depending upon the sing of $x_2$ we have that $\sinh(nx_2)/n^2 \to \pm \infty$. – user Nov 11 '18 at 20:18
  • ${\sin nx_1\mid n\in\Bbb{N}}$ is not dense in $[-1,1]$, if $x_1$ happens to be a rational multiple of $\pi$. – Jyrki Lahtonen Nov 11 '18 at 20:24
  • @JyrkiLahtonen I've indeed considered $x_1 \neq k\pi$ with $k\in \mathbb{Z}$. Isn't it correct? – user Nov 11 '18 at 20:26
  • Check what happens with, say $x_1=\pi/5$. Only finitely many values of $\sin nx_1$ in that case. Doesn't look like a dense set. Doesn't really disturb the conclusion, just the justification. – Jyrki Lahtonen Nov 11 '18 at 20:28
  • @JyrkiLahtonen, can you please explain when we don't have a limit and how to prove that please ? Thanks – mostafa Nov 11 '18 at 20:31
  • @JyrkiLahtonen Thanks I get your point. I try to fix that. – user Nov 11 '18 at 20:32
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In general the limit does not exist. For example, if $x_1=\frac{\pi}{2}$ and $x_2=1$, the result is $n^{-2}\sinh n$ times either $0$ or $\pm 1$ depending on $n(\operatorname{mod} 4)$. This result is a vanishing subsequence, but also subsequences diverging to $\pm\infty$. On the other hand, the limit can exist, e.g. if some $x_i=0$.

J.G.
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