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Infinity limit of e^(-x) and a lot of Fluff

  1. Nov 8, 2014 #1

    RJLiberator

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    • Member warned about not using the homework template
    math_zps9d900b98.jpg

    So this goes to 0.

    Is this because e^(-x) going to infinity is = to 0 and thus both parts of this equation is equal to 0 ?
    If so, is there anyway I can prove this through my work?

    Thank you.
     
  2. jcsd
  3. Nov 8, 2014 #2

    LCKurtz

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    Take the absolute value of that expression, overestimate the sines and cosines and see what happens.
     
  4. Nov 8, 2014 #3

    RJLiberator

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    Overestimate the sin/cosines? I guess this is the part of the problem that is throwing me off (it seems).

    I know the limit as they go to infinity does not exist. However, with e^-x there, it seems like the limit must go to 0 as e^-x grows faster.
     
  5. Nov 8, 2014 #4
    Use the squeeze theorem. i.e. find another function [itex]g[/itex] that is greater than or equal to your function for positive x values and [itex]\lim_{x \to +\infty} g(x) = 0[/itex].
     
  6. Nov 8, 2014 #5

    LCKurtz

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    Use the triangle inequality ##|a+b| \le |a| + |b|## and ##|sin\theta| \le 1## etc.
     
  7. Nov 8, 2014 #6

    RJLiberator

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    AH. There it is LCKurtz. The hint that I needed to complete this problem.

    In understanding the triangle inequality, this proves that sin(x) can never be greater than 1 (and likewise cos(x)). Therefore the meager numbers of -1 to 1 go nowhere while the e^-x to infinity is the dominating force.

    Thank you.
     
  8. Nov 8, 2014 #7

    LCKurtz

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    But you need to write out the string of inequalities to solve your problem.
     
  9. Nov 8, 2014 #8

    RJLiberator

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    |sinθ|≤1
    |cosθ|≤1

    Therefore
    |cosθ|+|sinθ|≤2

    And then I just take the limit of both parts separately due to the limit rules.
     
  10. Nov 8, 2014 #9

    LCKurtz

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    That isn't enough. You have to start with the absolute value of your original expression which, unfortunately, you posted as an image. Then with a string of inequalities you end up with something you can show goes to zero. You are going to have to type that expression if you are going to post an argument here.
     
  11. Nov 8, 2014 #10

    Ray Vickson

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    Your function is of the form ##f(x) = e^{-x} g(x)##, where ##g(x)## is a bounded function.
     
  12. Nov 8, 2014 #11

    LCKurtz

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    which he has not yet written the inequalities to prove...
     
  13. Nov 8, 2014 #12

    RJLiberator

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    Ok, so essentially, by taking the e^-x out we are left with:

    e^-x*(((cos(x)+sin(x))/2)-(cos(x)+1)).

    Now, the two elements in the parenthesis cannot be greater than 2. This is clear to me under any values of x.

    By using the triangle inequality I can prove that the inside values cannot be greater then what the triangle inequality states. Correct so far?
     
  14. Nov 9, 2014 #13

    LCKurtz

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    You seem to have an aversion to actually writing out the inequalities. I know you understand the ideas but I'm not going to say it's correct until I see the inequalities correctly written out.
     
  15. Nov 9, 2014 #14

    RJLiberator

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    Fair enough. Let me try my hand at this:

    1. We can factor an e^-x out:

    Limit as x goes to infinity of e^-x (sin(x)+cos(x)/2-(cos(x)+1))

    2. We can now make inequalities

    (sin(x)+cos(x))/2 cannot be greater than 1
    cos(x)+1 cannot be greater than 2

    With e^-x taking the equation to 3/infinity this means the limit goes to 0.

    I don't think it was an aversion to writing out the inequalities as much as not knowing 'exactly' what to write :/.

    I get the general idea, but is this the correct way to present it?
     
  16. Nov 9, 2014 #15

    LCKurtz

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    No, it isn't the correct way to present it. Forget the limit as ##x\to \infty## for a moment. Just start with$$
    \left |e^{-x}\left(\frac{ (\sin x + \cos x)}{2} - (\cos x + 1)\right)\right | \le$$On the right side of this inequality you should put an expression that is greater than the left side. It will involve using the triangle inequality and have absolute value signs in it. Then continue overestimating.
     
  17. Nov 10, 2014 #16

    RJLiberator

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    $$
    \left |e^{-x}\left(\frac{ (\sin x + \cos x)}{2} - (\cos x + 1)\right)\right | \le$$

    |e^-x*(sin(x)+cos(x))/2| + |e^-x(cos(x)+1)|
    It seems fairly obvious what the triangle inequality tells us to do, the issue is what to do with the e^-x? I decided to factor it back in, but that doesn't seem right.
     
  18. Nov 10, 2014 #17

    HallsofIvy

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    I haven't read through this entire discussion, so this may already have been said, but looking at the original post, it seems to me that that the first thing I would do is note that, since sine and cosine are never larger then 1 nor less than -1, (sin(x)+ cos(x))/2 is never larger than 1 nor less than -1 (since sine and cosine are never 1 at the same time you can get a "tighter" bound, but this is sufficient) and cos(x)+ 1 is never larger than 2 so this whole thing is never larger than [itex]3e^{-x}[/itex] nor less than [itex]-3e^{-x}[/itex] and those are sufficient to show that the limit, as x goes to 0, is 0.
     
  19. Nov 10, 2014 #18

    LCKurtz

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    You eventually want ##e^{-x}M## where ##M## is some fixed positive bound so you can say it goes to zero as ##x\to \infty##. Keep applying the triangle inequality and use the fact that ##|ab| = |a||b|## and eventually use ##|\cos x|\le 1 ## etc.
     
  20. Nov 10, 2014 #19

    LCKurtz

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    Perhaps you should have read the discussion before posting and you would have seen that I am trying to show the OP how to make a real argument out of that heuristic discussion.
     
  21. Nov 14, 2014 #20

    RJLiberator

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    Hey guys,

    I just wanted to update you on this thread:

    I appreciate your help and assistance with this problem. If I had more time during the week to solve it, I would have likely researched more. Unfortunately, with projects/exams in other courses I didn't get as in depth as I would have wished.

    LCK, I thank you for your continued guidance in my threads and being patient with me.

    -Ron
     
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