Limit - which result is correct?

  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. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    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. Ok, so now I put the "L'Hospitaled" function to the Maple and it still gives me symbolic result...
     
  5. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

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

    Daniel.
     
  6. ehild

    ehild 12,457
    Homework Helper
    Gold Member
    2014 Award

    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. 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. HallsofIvy

    HallsofIvy 40,953
    Staff Emeritus
    Science Advisor

    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. I didn't want to prove the MAPLE calculation, I was trying to prove the teacher's calculation...
     
  10. nrqed

    nrqed 3,083
    Science Advisor
    Homework Helper
    Gold Member

    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. ehild

    ehild 12,457
    Homework Helper
    Gold Member
    2014 Award

    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. Nice ehild, I thought this is one of the limits which can't be solved without L'Hospital's rule...Thank you.
     
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?
Similar discussions for: Limit - which result is correct?
Loading...