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Limits as x approaches infinity

  1. May 29, 2014 #1
    Went to a tutor for help with this problem and he was stumped

    need to find the limit

    2 - t + sin t/ t + cos t

    book says answer is -1

    My attempts

    I tried two things
    1. I just substituted 0

    2-0 + sin(0)/ 0 + cos(0)
    2/1
    2 wrong answer

    2. I divided everything by the highest t
    (2/t - t/t + sin(t)/t)/(t/t+cos(t)/t)
    (2/t -1 +1)/(1 + cos(t)/t)
    2t/1+cos(t)/t I got stuck right here
     
  2. jcsd
  3. May 29, 2014 #2

    HallsofIvy

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    You titled this "Limits as x goes to infinity". But there is no "x" in what you have here. Did you mean "as t goes to infinity"?

    You mean (2- t+ sin t)/(t+ cos t)
    [tex]\frac{2- t+ sin(t)}{t+ cos(t)}[/tex]
    right?
    Divide both numerator and denominator by t:
    [tex]\frac{\frac{2}{t}- 1+ \frac{sin(t)}{t}}{1+ \frac{cos(t)}{t}}[/tex]
    Now, it should be clear that, as t goes to infinity, 2/t goes to 0. Also, because for any t, sin(t) and cos(t) lie between -1 and 1, as t goes to infinity, sin(t)/t and cos(t)/t also go to 0.

    Why "0"?? Is the limit as t goes to 0 or infinity?

    How did "sin(t)/t)" become "1"?

     
  4. May 29, 2014 #3

    micromass

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    Can you please please please please please please use brackets! Don't write what you wrote above, but write (2 - t + sin(t))/(t + cos(t)).
     
  5. May 29, 2014 #4
    Yes I did mean t

    I used 0 because an example in the book just substituted 0

    sin(t)/t became 1 because of the rule sinθ/θ = 1
     
  6. May 29, 2014 #5

    micromass

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    I don't know where you saw that rule, but it's definitely wrong.

    What you probably meant was

    [tex]\lim_{\theta\rightarrow 0}\frac{\sin(\theta)}{\theta} = 1[/tex]

    This is true but unhelpful here, since the above is the limit as ##x## approaches ##0##. You want the limit as ##x## approaches infinity.

    So you need to find

    [tex]\lim_{\theta\rightarrow +\infty}\frac{\sin(\theta)}{\theta}[/tex]

    Do you know the squeeze theorem?
     
  7. May 29, 2014 #6
    Yeah that was the rule I was talking about

    I know the squeeze theorem but don't fully understand it.
     
  8. May 29, 2014 #7
    So I substitute infinity

    [(2/∞)-1 + sin∞/∞]/[1+(cos∞/∞)]
    (0-1)/1
    -1/1
    -1
     
  9. May 29, 2014 #8

    Ray Vickson

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    You wrote your function ##f(t)## as
    [tex] f(t) = 2 - t + \frac{\sin(t)}{t} + \cos(t),[/tex]
    whose "limit" as ##t \to +\infty## is ##-\infty##, while the limit as ##t \to 0## is 3. On the other hand, if you meant
    [tex] f(t) = \frac{2 - t + \sin(t)}{t + \cos(t)} [/tex]
    the limit as ##t \to +\infty## is -1 while the limit as ##t \to 0## is 2.
    So, what is the problem, exactly? I cannot figure out anything that you are doing because of your lack of relevant parentheses.
     
  10. May 29, 2014 #9
    yeah sorry about that ray.

    I don't need help anymore thanks hallsofivy and micromass you two helped a lot.
     
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