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Need help with trig limit

  1. Aug 30, 2014 #1
    1. The problem statement, all variables and given/known data

    lim x->0 (2sinxcosx)/ (2x^2 + x )

    2. Relevant equations
    2sinxcosx = sin(2x)


    3. The attempt at a solution
    denom. factors to x(2x +1) how to proceed?
     
  2. jcsd
  3. Aug 30, 2014 #2

    Mark44

    Staff: Mentor

    Your expression can be rewritten as ##\frac{\sin(2x)}{2x(x + 1/2)}##. Does that help?
     
  4. Aug 30, 2014 #3
    the limit of sin2x/2x will be 2. am I on the right track?
     
  5. Aug 30, 2014 #4
    sin2x/2x will be 1 I mean
     
  6. Aug 30, 2014 #5
    can it be written as sin2x/2x * 1/(x + 1/2)
     
  7. Aug 30, 2014 #6

    PeroK

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    Science Advisor
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    Gold Member

    Yes, that's the idea.
     
  8. Aug 30, 2014 #7
    I get lim = 2
     
  9. Aug 30, 2014 #8
    very helpful
     
  10. Aug 30, 2014 #9

    PeroK

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    Yes, but note that you could actually have done it directly:

    [tex]\lim_{x→0}\frac{2sin(x)cos(x)}{2x^2+x}= 2\lim_{x→0}\frac{sin(x)}{x}cos(x)\frac{1}{2x+1}=2[/tex]
     
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