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Limits of trig functions

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data

    lim (cos x - 1) / (sin^2 x + x^3) as x approaches 0.

    2. Relevant equations

    sinx/x = 1

    3. The attempt at a solution
    I get 0/0. Is that the answer?
     
  2. jcsd
  3. Dec 5, 2009 #2

    Dick

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    Yes, it has 0/0 form. That doesn't mean the limit is 0. Do you know l'Hopital's rule?
     
  4. Dec 5, 2009 #3

    ideasrule

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    Just a note: most of the limits you do will be in the form 0/0. What the limit actually equals to can be anything from negative infinity to positive infinity.
     
  5. Dec 6, 2009 #4
    I know l hospitals rule u take the derivative and then find the limit

    but is there another way u can so it by using sinx/ x = 1
     
  6. Dec 6, 2009 #5

    Dick

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    Yes, there's another way. Start by multiplying numerator and denominator by cos(x)+1 and expand the numerator. Try it and see how far you get.
     
  7. Dec 6, 2009 #6
    cos^2 - 1 / ( sin^2 + x^3)(cos x + 1)
    = - sin^2/ (sin^2 + x^3 )(cos +1)

    is this the right way?
     
  8. Dec 6, 2009 #7
    i still get 0/0
     
  9. Dec 6, 2009 #8

    Dick

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    You are doing fine. Now divide numerator and denominator by sin(x)^2.
     
  10. Dec 6, 2009 #9
    - 1/cos + 1 + sin^2/x^3 cos x + sin^2/X^3
    limit = 1?
     
  11. Dec 6, 2009 #10

    Dick

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    You already weren't using enough parentheses to make it clear what you mean. Now you've lost all of them. I have no idea what you are trying to write. Start over and show the algebra steps you are using.
     
  12. Dec 7, 2009 #11
    (1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3)

    limit =1
     
  13. Dec 7, 2009 #12
  14. Dec 7, 2009 #13

    Dick

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    Start again from -sin(x)^2/((sin(x)^2+x^3)(cos(x)+1)). And explain step by step you got there. Divide numerator and denominator by sin(x)^2. There's no need to multiply the denominator out. What's lim x->0 of cos(x)+1?
     
  15. Dec 7, 2009 #14
    Whs^

    BTW I cut and pasted a png into my blog and got warned for it as a Homework Q?

    Is that really necessary?

    https://www.physicsforums.com/blog.php?b=1522 [Broken]

    ie and who should I ask about it?

    No offence but I don't think trig identities warrant that much attention do they?
     
    Last edited by a moderator: May 4, 2017
  16. Dec 7, 2009 #15

    Dick

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    No idea. I can't see the blog entry anyway. You should take it up with the mentor who issued the warning. BTW, I don't see what sigmoids have to do with this problem. It's just a simple limits exercise.
     
    Last edited by a moderator: May 4, 2017
  17. Dec 7, 2009 #16
    Sure np thanks Dick, helpful as ever and quick.

    And congrats on being made a mod/homeworkhelper/Mentor btw, well deserved.
     
  18. Dec 7, 2009 #17
    1+ (sin^2/x^3(cosx+1))

    limit of cos + 1 = 2
     
  19. Dec 7, 2009 #18

    Dick

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    Yes, the limit of cos(x)+1 is 2. As for "1+ (sin^2/x^3(cosx+1))", if that's supposed to be the denominator, it's still messed up. If you aren't going to show your algebra steps, I really can't tell you how you are messing up.
     
  20. Dec 7, 2009 #19
    -sin^2 / (sin^2 + X^3(cos +1))
    = 1/(1+ (x^3(cos +1))/sin^2)
    lim= 1
     
  21. Dec 7, 2009 #20

    Dick

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    That's -sin(x)^2 / ( (sin(x)^2+x^3)*(cos(x)+1) ). (cos(x)+1) multiplies both sin(x)^2 and x^3, not just one of them. So you get -1/( (1+x^3/sin(x)^2)*(1+cos(x)) ). Now what's the limit of x^3/sin(x)^2?
     
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