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Homework Help: Can someone check this limit please

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    find the limit of:


    3. The attempt at a solution

  2. jcsd
  3. Feb 22, 2012 #2


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    Homework Helper

    if you're talking about using taylor expansions, then you should do the same thing for tan(x) to show its behaviour in the limit, and your expnasion is not correct due to the 2 in cos(2x)

    however as this is 0/0 indeterminate I think you should use L'Hopital (which is in essence a similar thing)
    Last edited: Feb 22, 2012
  4. Feb 22, 2012 #3
    you are doing right
  5. Feb 22, 2012 #4


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    Staff Emeritus
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    That should be
    [itex]\displaystyle \frac{1-\cos(2x)}{\tan(x)}=\frac{1-\left(1-\frac{(2x)^{2}}{2}+\dots\right)}{\tan(x)}=\frac{1-1+\frac{(2x)^{2}}{2}-\dots}{\tan(x)}[/itex]​

    You can do this without using L'Hôpital's rule or Taylor series.

    Start with cos(2x) = 1 - 2sin2(x) and 1/tan(x) = cos(x)/sin(x) .
  6. Feb 22, 2012 #5


    Staff: Mentor

    The lim symbol should remain until you actually take the limit. In the expression in the middle, above, the denominator is 0, so you can't conclude that the limit is zero.
    See above.
  7. Feb 22, 2012 #6


    Staff: Mentor

    This is definitely the way to go.
  8. Feb 22, 2012 #7
    after doing that i got:


    Is that correct?
  9. Feb 22, 2012 #8


    Staff: Mentor

    No. How did you get from (1 - cos(2x))/tan(x) to (-sin2(x)cos(x))/sin(x)? Show us what you did to get that.

    Also, the lim symbol needs to appear until you actually take the limit. In other words, as long as x is still in any expression, you haven't takent the limit.
  10. Feb 22, 2012 #9
    Well, i replaces 1-cos(2x) with 1-1-sin^2(x) and the bottom from tan(x) to sin(x)/cos(x)

    Then i removed the denominator by 1/tan(x).
  11. Feb 22, 2012 #10


    Staff: Mentor

    You have two mistakes. That should be 1 - (1 - 2sin2(x)).
  12. Feb 22, 2012 #11
    alright and for the denominator should i do tanx=sinx/cosx?
  13. Feb 22, 2012 #12
    Ok i got (2sin^2(x)*cos(x))/(sin(x))

    then i canceled out with the sine and i was left with 2sinx*cosx which then equals 0 is that correct?
  14. Feb 22, 2012 #13


    Staff: Mentor

    2sin(x)cos(x) doesn't equal 0, but the limit as x approaches 0 of this expression is 0.

    In other words,
    2sin(x)cos(x) ≠ 0, but [itex]\lim_{x \to 0} 2sin(x)cos(x) = 0[/itex]. There's a difference here.
  15. Feb 22, 2012 #14
    Yea sorry i didn't write, that but is what i did correct?
    Last edited: Feb 22, 2012
  16. Feb 22, 2012 #15


    Staff: Mentor

    The value of the limit is 0, but what you wrote is not correct.
    Again, 2sinx * cosx ≠ 0, in general.
  17. Feb 22, 2012 #16
    I don't get it what's wrong with that. Isn't the limit as x approaches 0 of 2sinx*cosx=0?
  18. Feb 22, 2012 #17


    Staff: Mentor

    Yes, but what you said didn't include the word "limit," so isn't true.

    [tex]\lim_{x \to 0}~2sin(x)cos(x) = 0[/tex]
    2sin(x)cos(x) = 0

    I'm am trying to get you to distinguish between the value of an expression and the value of the limit of that expression.

    Do you see the difference now?
  19. Feb 22, 2012 #18
    Yes thank you I appreciate it :smile:.
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