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Limit - which result is correct?

  1. Jan 15, 2005 #1
    Hi all,

    I've been practising for the exam and did some limits from our master's collections:

    [tex]
    \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan^{3} x - 3\tan x}{\cos \left( x + \frac{\pi}{6} \right) }
    [/tex]

    I got the result -24, but there is written -12. Which one is correct? I tried to prove the result in Maple and Mathematica, but in neither I'm not able to get anything but symbolic result.

    Thank you.
     
  2. jcsd
  3. Jan 15, 2005 #2

    dextercioby

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    HINT:Bring your function to anither form,so that the limit can be put under the "conventional" form "->0".That requires a change of variable.

    Daniel.

    P.S.Can as in do you know how to apply L'Hôspital's rule??
     
  4. Jan 15, 2005 #3
    Ok, so now I put the "L'Hospitaled" function to the Maple and it still gives me symbolic result...
     
  5. Jan 15, 2005 #4

    dextercioby

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    Can Maple compute limits??Can u compute derivatives??CAN U BOTH SOLVE THIS SIMPLE PROBLEM?????

    Daniel.
     
  6. Jan 15, 2005 #5

    ehild

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    Your result is correct.

    Just try to choose some value for x close enough to pi/3 and calculate the replacement value. You will see that it is far away from -12.

    ehild
     
  7. Jan 15, 2005 #6
    I don't understand. Using L'Hospital I got the result I wrote here. But the "official" result is different to mine. I was just asking how to prove it in Maple.
     
  8. Jan 15, 2005 #7

    HallsofIvy

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    The point was: Why use MAPLE?? You can't PROVE the MAPLE calculation is correct by using MAPLE!
    It's easy to calculate that the derivative of
    tan3(x)- 3tan(x) is 3tan(x)sec2(x)- 3sec2(x) and certainly you should know that [itex]tan(\frac{\pi}{3})= \sqrt{3}[/itex] and that [itex]sec(\frac{\pi}{3}})= 2[/itex]. Of course, the derivative of [itex]cos(x+\frac{\pi}{6}[/itex] is [itex]-sin(x+\frac{\pi}{6})[/itex] and [itex]-sin(\frac{\pi}{2}})= -1[/itex].

    The limit, by L'Hopital, is (3*3*4- 3(22)/-1= -(36-12)= -24.
     
  9. Jan 16, 2005 #8
    I didn't want to prove the MAPLE calculation, I was trying to prove the teacher's calculation...
     
  10. Jan 16, 2005 #9

    nrqed

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    That's indeed correct (notice that there's a typo, it should obviously be

    "It's easy to calculate that the derivative of
    tan3(x)- 3tan(x) is 3tan2(x)sec2(x)- 3sec2(x)"

    i.e. there is a square in the first tan in the derivative but it's obviously a typo because HallsofIvy gave the correct numerical answer.

    Just a thought: if someone forgets to square the sec(x), they will get -12. So that could be the origin of the incorrect answer -12.

    Pat
     
  11. Jan 16, 2005 #10

    ehild

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    The limit can be obtained even without L'Hopital's rule.

    [tex]
    \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan^{3} x - 3\tan x}{\cos \left( x + \frac{\pi}{6} \right) }=
    \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan(x)(\tan^{2} x - 3)}{\cos ( x )\cos (\pi/6)-\sin(x)\sin(\pi/6)}=[/tex]

    [tex]\lim_{x \rightarrow
    \frac{\pi}{3}}\frac{\tan(x)(tan(x)-\sqrt(3))(\tan(x)+\sqrt(3))}{1/2\cos(x)(\sqrt(3)-\tan(x))}=
    \lim_{x \rightarrow \frac{\pi}{3}}\frac{-2\tan(x)(\tan(x)+\sqrt(3))}{\cos(x)}=\frac{-2\sqrt(3)(2\sqrt(3)}{1/2}=-24
    [/tex]

    ehild
     
  12. Jan 16, 2005 #11
    Nice ehild, I thought this is one of the limits which can't be solved without L'Hospital's rule...Thank you.
     
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