Is it correct to say that a decaying oscillation goes to zero?

Question

Let's say we have a function $f(x)= e^{-2x} (sin(x)+ cos(x))$, and we want to describe its behavior as $x \to \infty$ . We know that this function is a decaying oscillation, but is it mathematically correct to say that as $x \to \infty , f \to 0$ ? I think it is wrong to say this because the limit as $x \to \infty$ does not exist, (the function oscillates and does not remain above/below the x-axis).

Answer 1

Yes it is correct to say that $$f(x)= e^{-2x} (sin(x)+ cos(x))$$ approaches to $0$ as $x\to \infty.$

It is mathematically correct because $$|e^{-2x} (sin(x)+ cos(x))| \le 2 e^{-2x} \to 0$$

Answer 2

It is correct to say as $x \to \infty$ that $f \to 0$, because we can squeeze the function $f(x)= e^{-2x} (sin(x)+ cos(x))$ using the two functions $2e^{-2x}$ and $-2e^{-2x}$, and since the limit of both $2e^{-2x}$ and $-2e^{-2x}$ as $x \to \infty$ is 0, then the squeeze theorem says that the limit of $e^{-2x} (sin(x)+ cos(x))$ as $x \to \infty$ is 0.