# 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.?

• Hint: Set $a_n=-2n\pi$ for $n=0,1,2, \dots$ Sep 26 '20 at 13:38
• It oscillates forever. It's also not bounded, but it doesn't blow up to either infinity in particular.
– Ian
Sep 26 '20 at 13:38

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.

• Thank you. Why is it sufficient to show that the sequence approaches different infinities to prove the limit does not exist? Sep 26 '20 at 13:56
• @aiyan For the uniqueness theorem of the limit, when limit exists it is unique therefore if we find at least $2$ subsequances with different limits the limit doesn't exist.
– user
Sep 26 '20 at 14:00

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.

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