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Homework Help: Question on a limit rule

  1. Dec 9, 2007 #1
    [SOLVED] Question on a limit rule

    1. The problem statement, all variables and given/known data
    Question 3. My question is (Using t as theta, as I'm too lazy to use TeX), when going from the limit of sin(3t)cos(5t)/sin(5t), the person then breaks up this limit into 3 seperate limits (All multiplied by each other), then reduces those 3 limits down to 3/5, 1, and 1.

    Where did those 3 limits come from? What is the rule or theorem that I'm missing here?
    I get how they break down into 3/5, 1, and 1, but I don't see how he split the first into those 3.

    2. Relevant equations

    3. The attempt at a solution
    Google? :P
  2. jcsd
  3. Dec 9, 2007 #2


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    They've just multiplied the top and bottom of the expression by 3, 5 and theta. That is, they have multiplied the expression by 1, three times.
  4. Dec 9, 2007 #3


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    First of all, there's a mistake in the solution; in the first of the three limits, the cos(5t) should be in the numerator, not the denomator.

    Correcting this, the product of the three expressions (before you take the limit), equals the original expression, so this breakup is mathematically correct.

    Why choose this particular form? Because we know sin(x)/x -> 1 as x->0, so it's useful to put in a factor of 1/x for each sin(x) (in the numerator or denominator). The leftover stuff then turns out be a number times cos(x), and the limit of this is easy as well. So the general idea was to write the original expression as a product of expressions whose limits are well known.
  5. Dec 9, 2007 #4
    At first I thought I got it, but then something else threw me off.

    Without multiplying by 1, sin(3t)cos(5t)/sin(5t) would break into:
    Limit of [ sin(3t) / 1 ]
    Limit of [ cos(5t) / 1 ]
    Limit of [ 1 / sin(5t) ]

    However their three limits have one sin on the top, one sin on the bottom, and one cos on the bottom.

    How did their cos get to the bottom of the fraction?
  6. Dec 9, 2007 #5


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    Yea, sorry, I didn't notice that; see the above post by Avodyne.
  7. Dec 9, 2007 #6
    Good to know. Thank-you for your help :)
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