Finding a limit of a function of two variables

Question

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.

Answer 1

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

• Find all $x$ such that lim $\sin (nx)$ exists

Answer 2

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 $x_i=0$.