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HELP! Another limit question!

  1. Oct 20, 2004 #1
    Hi, I have this question that I am not all to sure how to do. Actually, I don't know where to even start.

    [itex]\large r(x)=\frac{3x^2 - Sx - S^2}{2x^2 + Sx + S ^ 2}[/itex] Where S represents some abitrary number, let's say 9 for an example.

    I found the limit of the function to be [itex]\frac{3x}{2}[/itex]

    Then the question asks me to find a value of N that satisfies [itex]|r(x) - \frac{3x}{2}| < 0.001[/itex]

    I have no clue of where to start... I have these examples but they didn't help much.
     
  2. jcsd
  3. Oct 21, 2004 #2

    Galileo

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    So you're taking the limit of r(x) as x goes to ???

    Furthermore, the limit (if it exists) is a number. Your limit has an x-dependence.

    I assume you take the limit as x goes to infinity, since that would resemble your answer closely.
    [tex]\lim_{x \rightarrow \infty}r(x)=\frac{3}{2}[/tex]

    Then I guess you must have [itex]|r(x)-\frac{3}{2}|<0.001[/itex] whenever [itex]x>N[/itex] for some number N.
    I think N will depends on S. If S=9, you can check that N=1.000.000 for example easily does the job.
     
  4. Oct 21, 2004 #3

    arildno

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    Write:
    [tex]|r(x)-\frac{3}{2}|=|\frac{3x^{2}-Sx-S^{2}}{2x^{2}+Sx+s^{2}}-3/2\frac{x^{2}+Sx+S^{2}}{x^{2}+Sx+S^{2}}|=[/tex]
    [tex]\frac{2}{x^{2}}|\frac{\frac{S}{x}+(\frac{S}{x})^{2}}{1+\frac{1}{2}(\frac{S}{x}+(\frac{S}{x})^{2})}|[/tex]
    1. Assume that x is so big that:
    [tex]|\frac{S}{x}+(\frac{S}{x})^{2}|<1[/tex]
    Hence, we have:
    [tex]|r(x)-\frac{3}{2}|\leq\frac{2}{x^{2}}\frac{1}{1-\frac{1}{2}}=(\frac{2}{x})^{2}[/tex]
    2. From this it is simple to find a value of x which gives you your estimate.
    3. You are not quite finished, though:
    Evidently, you must find an N so that BOTH your inequalities,
    [tex]|\frac{S}{x}+(\frac{S}{x})^{2}|<1[/tex], [tex](\frac{2}{x})^{2}<\epsilon[/tex]
    are fulfilled, i.e, choose your N to be the maximal value of the "separate N's" you derive for each inequality.
    Your final N is therefore dependent on both [tex]S,\epsilon[/tex]
    ([tex]\epsilon[/tex] is the stated error margin)
     
    Last edited: Oct 21, 2004
  5. Oct 21, 2004 #4
    Thanks, I don't have time to look at this yet. I'm late for class. But the limit was as [itex]x[/itex] approaches [itex]infinity[/itex]. Sorry I should of mentioend that
     
  6. Oct 21, 2004 #5

    arildno

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    Sure, I know that the limit goes to infinity; the derived inequalities holds for "the absolute value of x greater than some number"
     
    Last edited: Oct 21, 2004
  7. Oct 21, 2004 #6

    Sorry, the message was for the post prior to yours... the one by... Galileo. Thanks for you help! I still don't have time to look at this... I'll give it a gander when I get home.
     
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