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I have calculus homework question that to be quite frank I don't even know how to begin to solve. Here is my attempt on how to to solve the question; however I'm not sure if I'm tackling this correctly. I'm hoping that someone here can tell me if I'm doing this correctly and if my answer would be correct.

Here is the question.

Find the limit or show that it does not exist: $$\lim_{x\to \infty}(e^{-x}+2\cos3x)$$

My attempt to the solution is DNE or does not exist simply because cosine oscillates between -1 and 1 and it never approaches any one number to reach a horizontal asymptote. Is this answer correct? Or how should I approach this problem if it's not? Or should it be $$-\infty$$ since that would be the the value of $$e^{-x}\quad ?$$Please forgive my stupidity if I'm incorrect in all fronts.

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This is not a bad question. On the contrary, you show you have put some real thought into it. Good work, Miguel. – ncmathsadist Feb 10 '13 at 1:38
Miguel, I want to commend you for explaining your thoughts and work on the problem when asking your question. All too often, people just post the statement of the problem and wait for someone to do their work. Demonstrating that you've tried the problem yourself makes people more willing to help you, and also enables people to understand possible misunderstandings you have, so they can help you correct them. This is a fantastic question! – Zev Chonoles Feb 10 '13 at 1:38
"Should it be $-\infty$ since that would be the value of $\mathrm{e}^{-x}$?" I don't think the value of $\mathrm{e}^{-x}$ approaches $-\infty$, if that's what you meant. – Herng Yi Feb 10 '13 at 1:48
@ncmathsadist and @ Zev Chonoles thank you guys I appreciated Im a believer thats i should learn how to fish on my own if you know what i mean. Thanks for the kind words. – Miguel Feb 10 '13 at 2:00
up vote 1 down vote accepted

You're correct that the limit does not exist:

$$\lim_{x\to \infty}(e^{-x}+2\cos3x) = \lim_{x\to \infty} \left(\frac{1}{e^{x}} + 2 \cos 3x\right) =\lim_{x\to \infty} (0 + 2 \cos 3x)$$

This sum vacillates between $2^+$ and $-2^+$ as $x \to \infty$. Hence the limit does not exist.

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doesn't the limit distribute over the sum only when both sublimits are well-defined? I get what you mean, but what you wrote is not rigorous. – Herng Yi Feb 10 '13 at 1:45
@HerngYi I've "rephrased" the problem... – amWhy Feb 10 '13 at 2:03
Vacillates or oscillates? Which one is the best here? + – Babak S. Feb 10 '13 at 3:40
"Definition of VACILLATE. 1. a : to sway through lack of equilibrium. b : fluctuate, oscillate." I take it that they are synonymous. – amWhy Feb 10 '13 at 3:42

Two things you need to notice. The $e^{-x}$ term decays to zero as $x\to\infty$.
The other term oscillates between $2$ and $-2$. So the whole sum oscillates between a little more than $2$ and a little more than $-2$.

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Consider the sequence $(x_n)$ where $x_n = \mathrm{e}^{-n\pi/3} + 2\cos(3(n\pi/3)) = \mathrm{e}^{-n\pi/3} + 2(-1)^n$. Note that $\lim_{i \to \infty}x_{2i} = 2$ and $\lim_{i \to \infty}x_{2i + 1} = -2$. Since $(x_n)$ has two distinct limit points, it cannot be convergent, and the limit in the question cannot exist.

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Thank you you answer further explains what i was trying to understand. – Miguel Feb 10 '13 at 2:03

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