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Homework Help: Basic trig manipulation in limits

  1. Jun 8, 2010 #1
    Hi

    Given this limit
    [tex]\lim_{h \to 0} \frac{1-cos(h)}{h} = 0[/tex]

    can you rearrange

    [tex](sin(x) (\frac{cos(h)-1}{h}))[/tex]

    to

    [tex](sin(x) \cdot -1 \cdot (\frac{1-cos(h)}{h}))[/tex]

    so that when you take the limit as h goes to 0 you get: sin(x) x -1 x 0 = 0

    thanks
     
  2. jcsd
  3. Jun 8, 2010 #2
    Also, how do i get 'big brackets' when i write my equations

    many thanks
     
  4. Jun 8, 2010 #3

    Mark44

    Staff: Mentor

    [tex]\lim_{h \to 0} sin(x) \left(\frac{1-cos(h)}{h}\right) = sin(x) \lim_{h \to 0} \left(\frac{1-cos(h)}{h}\right)= sin(x) \cdot 0 = 0[/tex]

    sin(x) doesn't depend on h, so as far as the limit process is concerned, sin(x) is a constant. The limit of a constant times a function is the constant times the limit of the function.

    For big parentheses use \left( and \right) inside your tex tags. For big brackets use \left[ and \right]
     
  5. Jun 8, 2010 #4
    Thanks

    but the equation is in the form (cos(h)-1)/h not 1-cos(h)/h

    that's why i was asking about multiplying by -1 to get it in the right form
     
  6. Jun 8, 2010 #5
    This should answer your question. In this case, it matters not since the limit is 0, but in general a change of sign is in order.
     
  7. Jun 8, 2010 #6

    Mark44

    Staff: Mentor

    Instead of multiplying by -1, factor out -1.

    (cos(h) - 1)/h = -1(-cos(h) + 1)/h = -(1 - cos(h))/h

    As Tedjn points out, it doesn't make any difference in this case, since the limit of this expression is 0.
     
  8. Jun 8, 2010 #7
    Sorry i meant factor out -1, not multiply

    I'm teaching this myself and not very confident which is why i wanted to check why the 2 forms both have a limit of 0


    I know how to show that the limit of the first form (1-cos(h)/h) = 0 but the only way i know how to show the limit of the second form (cos(h)-1/(h) = 0 is by factoring out the -1 to get the first form. Is it possible to show that the limit of the second form is 0 directly?

    With the first form you multiply the numerator by its conjugate, 1+cos(x), to get
    1-cos(x) . 1+cos(x) all over
    x . 1 + cos(x)

    this becomes
    sin2(x)/x . 1/(1+cos(x)) to get sin(x) . sin(x)/x . 1/(1+cos(x))
     
  9. Jun 8, 2010 #8
    In the second, you get a first term -sin(x) instead; that is the only difference. The conclusion then follows for the same reason. It's because the -1 carries through the entire way.
     
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