Find $\lim\limits_{x\to -\infty} (e^{-x} \cos{x})$ $$\lim\limits_{x\to -\infty} (e^{-x} \cos{x})=\lim\limits_{x\to -\infty} \left(\dfrac{\cos{x}}{e^x}\right)$$
From there, I see that $e^x$ approaches $0$ while $\cos{x}$ oscillates between $-1$ and $1$.
My answer is that the limit does not exist. What is the proper reasoning to explain this? Does the limit oscillate forever, approach $\pm\infty$, etc.?

Yes your idea is right, to show that in a rigorous way let consider that for $x_n= -2\pi n \to -\infty$ as $n\to \infty$
$$e^{-x_n} \cos{x_n}=e^{-x_n}=\infty$$
and for $x_n= -\pi n$
$$e^{-x_n} \cos{x_n}=-e^{-x_n}=-\infty$$
therefore the limit doesn't exist.
