Prove that $ $ \limsup_{x\to\infty}\left(\cos x + \sin\left(\sqrt2 x\right)\right) = 2 $ $

Pretty much always when I ask a question here I do provide some trials of mine to give some background. Unfortunately, this one is such a tough one for me that I don’t even see a starting point.

Here are some observations though. Let’s denote the function under the limsup as $ f(x)$ : $ $ f(x) = \cos x + \sin\left(\sqrt2 x\right) $ $

Since sin’s argument contains an irrational multiplier the function itself is not periodic, perhaps this may be used somehow. I’ve tried assuming that there exists $ x$ such that the equality holds, namely: $ $ \cos x + \sin\left(\sqrt2 x\right) =2 $ $

Unfortunately, I was not able to solve it for $ x$ . I’ve then tried to use Mathematica for a numeric solution, but NSolve didn’t output anything in Reals.

The problem becomes even harder since there are some constraints on the tools to be used. It is given at the end of the chapter on “Limit of a function”. Before the definition of derivatives, so the author assumes the statement might be proven using more or less elementary methods.

Also, I was thinking that it could be possible to consider $ f(n),\ n\in\Bbb N$ rather than $ x\in\Bbb R$ and use the fact that $ \sin(n)$ and $ \cos(n)$ are dense in $ [-1, 1]$ . But not sure how that may help.

What would the argument be to prove the statement in the problem section?