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Homework Help: Limit of a long trig function.

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    x->0 ((sin5x)/(sin2x) - (sin3x)/(4x))

    2. Relevant equations
    i guess sinx/x as x approaches zero is 1.

    3. The attempt at a solution
    lets be honest, this is a simple question. i am getting a final answer of 4, but it is wrong apperently according to the book's solution. i want to see if anyone else got this answer

    I separated the limit into two limits, divided by 5x's and 2x's on the left limit and divided by 3x's on the right limit. this shouldnt be a problem (and yes i divided top and bottom by the same amounts so they would cancel out; thats not my mistake). please someone help
  2. jcsd
  3. Nov 2, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    You've obviously made an algebra error somewhere, If you show your work, we can tell you where your error is.
  4. Nov 2, 2008 #3


    Staff: Mentor

    Are you sure you have given us the problem as it appears in your book or assignment? Could it be lim sin(5x)/(2x) - sin(3x)/(4x)? If so, that's a much simpler problem, whose limit is 1.75, as x approaches 0.
  5. Nov 2, 2008 #4


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    Science Advisor

    [tex]\frac{sin(3x)}{4x}= \frac{3}{4}\frac{sin(3x)}{3x}[/tex]

    [tex]\frac{sin(5x)}{sin(2x)}= \frac{5}{2}\frac{sin(5x)}{5x}\frac{sin(2x)}{2x}[/tex]
  6. Nov 2, 2008 #5
    HOW IS IT 1.75!!!!!!!!!!!!!!!!!!!! wheres the flaw in my logic.

    (sin5x)/sin2x = [((sin5x)/(5x*2x)]/[(sin2x)/(5x*2x)]
    = (5/2x)/(2/5x))
    = 25/4

    (sin3x)/4x = [(sin3x/3x)]/[(4x/3x)]
    = 3/(4x/3x)
    = 3/(4/3)
    = 9/4

    25/4 - 9/4 = 16/4 = 4 where is my flaw!!!???
  7. Nov 2, 2008 #6


    Staff: Mentor

    Halls, are you missing a division symbol in your second equation?
  8. Nov 2, 2008 #7


    Staff: Mentor

    How did you get (5/2x)/(2/5x)) from the expression above it? lim (as x [tex]\rightarrow[/tex] 0) of sin(5x) / (5x) is 1, not 5, if that's what you did.

    The expression on the right side of your first line is equal to
    [tex]\frac{sin(5x)}{5x} * \frac{5x}{2x} * \frac{2x}{sin(2x)}[/tex]
    As x [tex]\rightarrow[/tex] 0, this expression approaches 1 * 5/2 * 1 = 5/2. Similarly sin(3x)/(4x) approaches 3/4, so the difference approaches 7/4 = 1.75.

    Regarding my earlier question about whether the denominator of the first fraction should have been 2x rather than sin(2x), it doesn't matter. For x close to 0, sin(x) [tex]\approx[/tex] x, and sin(2x) [tex]\approx[/tex] 2x.

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