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Limits of form 0 x 0 question!

  1. May 4, 2009 #1
    1. Evaluate the limit or determine that it does not exist. (Not sure if I did this correctly or how to do it if I'm wrong)



    2. lim(x⇒π/2) tan^2(x) ln(sin(x))
    -In text form: the limit as x goes to pi over two, of tangent squared of x, times the natural log of sine of x.




    3. My attempt was to change tan^2[x] into sin^2[x]/cos^2[x], where the limit would then give me 0/0, I then used L'Hospital's rule to get the numerator (not sure if I did the derivative correctly with the product rule):
    2cos[2x]ln[sin(x)] + -csc^2[x]sin^2[x]
    and the denominator as -2cos[2x]
    This gave me -1/0 which would mean the limit is zero, and does exist.
     
  2. jcsd
  3. May 4, 2009 #2

    Dick

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    Ok, so the limit is sin(x)^2*ln(sin(x))/cos(x)^2, where the second factor is the 0/0 part. Just apply l'Hopital to that. What's the derivative of the numerator and the denominator?
     
  4. May 4, 2009 #3

    Mark44

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    You have made a number of mistakes here.
    tan^2(x) approaches infinity as x approaches pi/2. It is not of the form [0/0].
    A limit of the form -1/0 does not mean that the limit is zero, and BTW, zero does exist.

    In your original limit, tan^2(x) * ln(sin(x) is the indeterminate form [infinity * 0]. If you rewrite your limit as ln(sin(x))/cot^2(x), this is of the form [0/0], which you can use L'Hopital's Rule on. Remember when you're using L'Hopital's Rule, you are taking the derivatives separately of the numerator an denominator. It's not the same as the quotient rule for derivative.
     
  5. May 4, 2009 #4
    I did make a mistake in that -1/0 is zero, that is not true, I was thinking 0/-1. I think if I represented tan^2[x] as 1/cot^2[x] that would give me the form of 0/0 correct? Now I think I can do the derivative of both the top and bottom. My first attempt in changing tangent to sines and cosines would take me down a very un-fun and endless path of derivation.
     
  6. May 4, 2009 #5

    Mark44

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    Yes
    Note that using the technique Dick suggested is not a complicated as you're making it out to be. You only have to use L'H's rule on this part: ln(sin(x))/cos(x)^2, since the limit of the other part is 1.
     
  7. May 4, 2009 #6
    Alright, I used Dick's method and found the derivative of both the top and bottom which turn out to be:
    -csc^2[x] (divided by)
    -2cos[2x]
    So now I have the form 0/2 correct? Thus the limit is zero?
     
  8. May 4, 2009 #7

    Mark44

    Staff: Mentor

    No, that's not right.
    Starting with ln(sin(x))/cos^2(x), how did you get -csc^2(x)/(-2cos(2x))?

    Let's take it in parts:
    d/dx(ln(sin(x)) = ?
    d/dx(cos^2(x)) = ?
     
  9. May 4, 2009 #8
    True, I derived the numerator twice, which is how I got -csc^2(x). (My bad)
    After actually writing this down on a piece of paper:
    d/dx(ln(sin(x)) is simply cot(x)
    d/dx(cos^2(x)) is -2cos(x)sin(x)
    but shouldn't the denominator be:
    d/dx(cot^2(x)) ? Because tan^2(x) is 1/cot^2(x)
     
  10. May 4, 2009 #9

    diazona

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    No - as has already been said, the limit of the sin^2(x) part is 1. So you don't need to involve that in the computation with L'Hôpital's rule. And once you take out the sin^2(x) part, what you're left with is ln(sin(x)) in the numerator and cos^2(x) in the denominator.
     
  11. May 4, 2009 #10

    Dick

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    You can do it that way. I was just suggesting you split the sin^2(x) off since you know the limit of that is 1 and just concentrate on the ln(sin(x))/cos^2(x) part. You can really do it either way. But when you get to the l'Hopital's result change everything into sines and cosines and see what cancels.
     
  12. May 5, 2009 #11
    Ah... Ok now I see where you both are coming from. After doing all of that and cancelling sines and cosines as suggested, I got an anwer of -1/2.
     
  13. May 5, 2009 #12

    Dick

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    Right.
     
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