Find $\lim\limits_{x\to -\infty} (e^{-x} \cos{x})$

Question

$$\lim\limits_{x\to -\infty} (e^{-x} \cos{x})=\lim\limits_{x\to -\infty} \left(\dfrac{\cos{x}}{e^x}\right)$$

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.?

Answer 1

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.

Answer 2

Take $x=\frac{\pi}{2}+\pi k$, where $k$ is integer and $k\rightarrow-\infty$.

We see that $e^{-x}\cos{x}=0.$

In another hand, for $x=\pi k$ and $k\rightarrow-\infty$ we have $|e^{-x}\cos{x}|\rightarrow+\infty$.

Id est, the limit does not exist.

Answer 3

Apply the ratio test on the sequence $x=-2\pi n$.