1
$\begingroup$

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

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.?

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

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.

$\endgroup$
2
  • $\begingroup$ Thank you. Why is it sufficient to show that the sequence approaches different infinities to prove the limit does not exist? $\endgroup$
    – aiyan
    Sep 26 '20 at 13:56
  • 1
    $\begingroup$ @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. $\endgroup$
    – user
    Sep 26 '20 at 14:00
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.