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Tricky Limits

  1. Sep 7, 2006 #1
    Limits

    I'm doing some online practice work... and I can't find out how to get the following two limits.... I know they are not undefined.

    The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.


    The limit as x approaches 0 of (x + tan(4x)) / sin(9x)

    Any help would be appreciated.
     
  2. jcsd
  3. Sep 7, 2006 #2
  4. Sep 7, 2006 #3
    We're not doing derivitaves yet.. so it wouldnt make sense.. but I could try.
     
  5. Sep 7, 2006 #4
    Doesn't work.
     
  6. Sep 7, 2006 #5
    I'm doing some online practice work... and I can't find out how to get the following two limits.... I know they are not undefined.

    The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.


    The limit as x approaches 0 of (x + tan(4x)) / sin(9x)

    Any help would be appreciated.
     
  7. Sep 7, 2006 #6
    I am getting 16 as my answer. I first broke [tex]\sin(4x)^2[/tex] into [tex](2\sin(2x)*\cos(2x))^2[/tex] and then I further broke this down to [tex]16\sin(x)^2\cos(x)^2(\cos(x)^2-\sin(x)^2)^2[/tex]. Then I didived it by x*tan(x) from which I got:

    [tex]16*\frac{\sin(x)}{x}\cos(x)^3(\cos(x)^2-\sin(x)^2)[/tex]

    Then ... here is the muddy part ... I assumed that the limit of the products is the product of the limits. Then, since the limit of sinx(x)/x is 1, you end up with 16 as your answer after you evalute the rest of the limit at x=0.
     
  8. Sep 7, 2006 #7

    Hurkyl

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    Then find a reason why you can assume that. It's not hard.


    Incidentally, if you know (sin x)/x --> 1 as x --> 0, there's a much easier way to deal with sin (4x)...
     
  9. Sep 7, 2006 #8
    i also got 16 for the first one using L'Hopitals rule (although I had to go to second derivatives) but I just expanded sin^2(4x) as [1-cos(8x)]/2 and did l'hopitals rule twice (making sure I kept my fractions seperate)
     
  10. Sep 7, 2006 #9

    Hurkyl

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    You're all making these way too hard. Think about it -- what is the intuitive meaning of:

    [tex]
    \lim_{x \rightarrow 0} \frac{\sin x}{x}
    [/tex]

    ? Does it tell you anything interesting geometrically or algebraically?

    Highlight if you need to see the answer: (it tells you that, when x is small, sin x looks and acts very much like x)

    So what does that tell you about these limits?
     
    Last edited: Sep 7, 2006
  11. Sep 7, 2006 #10

    HallsofIvy

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    The simplest method would be L'Hopital's rule. Have you tried that? If you don't want to do that, you might separate it as
    [tex]4\frac{sin(4x)}{4x}\frac{sin(4x)}{tan(x)}[/tex]
    You should know the limit of [itex]4\frac{sin(4x)}{4x}[/itex] and you should be able to use the trig identity sin(4x)= 2sin(2x)cos(2x) to reduce
    [tex]\frac{sin(4x)}{tan(x)}[/tex]
     
  12. Sep 7, 2006 #11

    HallsofIvy

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    This same question was asked under "homework". I am going to combine the two threads.
     
    Last edited: Sep 7, 2006
  13. Sep 7, 2006 #12

    Hurkyl

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    You shouldn't need the trig identity. :wink: The only facts you need are the elementary limit properties, and that the limits of (sin x)/x and (tan x)/x are both 1, as x goes to zero.
     
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