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Homework Help: Finding the limit of trignometric functions

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the Limit: lim h->0 of ( h sin h)/ ( 1-cos h)

    2. Relevant equations

    lim h->0 of (sin theta)/ theta = 1
    lim h->0 of (cos(h)-1)/h=0

    3. The attempt at a solution
    lim h->0 (h/1-cos h)*(sin h) * (1+cos h/ 1 +cos h)

    = lim h->0 (h(1+cos h) * sin h)/(1-cos^2 h)

    = lim h->0 (h(1+cos h))/sin h

    That's how far I got, I have no idea how to continue.
    Please and thank you!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2010 #2

    Mark44

    Staff: Mentor

    You're almost there!
    [tex]=\lim_{h \to 0} \frac{h}{sin(h)}\cdot (1 + cos(h))[/tex]

    You can break up the limit of a product to the product of limits, provided that both limits exist. Hint: look at your relevant equations.
     
  4. Sep 29, 2010 #3
    Okay, so I split them up, but would the equation: lim h->0 of sin h/ h = 1 be the same as the lim h->0 of h/sin h?? how does that work?

    and therefore be: 1 * lim h->0 (1+ cos h)
    = 1 * 2 = 2?
     
  5. Sep 29, 2010 #4
    Yup, I get 2 also.

    If you know limh→0 h/sinh = 1 then you know that
    limh→0 (sinh/h)-1 = 1.

    But (sinh/h)-1 = 1 ⇒ (sinh/h)= 1


    Edit: You don’t know l’hopital’s rule right? If you do this is a rather round-about way to do this problem.
     
    Last edited: Sep 29, 2010
  6. Sep 29, 2010 #5

    Mark44

    Staff: Mentor

    Agreed.
    Not sure what the missing symbol is - implies?
    But sinh/h [itex]\neq[/itex] 1. The limit as h -> 0 is 1, though.
    But if the OP doesn't know L'Hopital's Rule, this is the way to go.
     
  7. Sep 29, 2010 #6
    Yah, I haven't learned L'Hopital's Rule yet

    But thanks guys!
     
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