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Math Help - Limits

  1. #1
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    Limits

    Hello!

    Here is my equation

    lim sin5x-3x
    x->0 3x

    I think that I'm supposed to separate the sin5x/3x and the -3x/3x but do i bring the limit in front of both or only the sin5x/3x? and would the -3x/3x just work out to be -1?
    Ahh a bit lost, any help would be greatly appreciated! Thanks in advance
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  2. #2
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    This is not an equation.

    \displaystyle \lim_{x\to 0} \left( \dfrac{sin(5x)-3x}{3x} \right) =  \lim_{x\to 0} \left( \dfrac{sin(5x)}{3x} - 1 \right)

    To complete it, you'll use this formula: \displaystyle \lim_{x\to 0} \dfrac{sin(ax)}{ax}=1.

    Can you rewrite \dfrac{sin(5x)}{3x} so that the formula could be applied ?
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  3. #3
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    Quote Originally Posted by wassup View Post
    Hello!

    Here is my equation

    lim sin5x-3x
    x->0 3x

    I think that I'm supposed to separate the sin5x/3x and the -3x/3x but do i bring the limit in front of both or only the sin5x/3x? and would the -3x/3x just work out to be -1?
    Ahh a bit lost, any help would be greatly appreciated! Thanks in advance
    Since this goes to \displaystyle \frac{0}{0}, L'Hospital's Rule is the quickest method of evaluating this limit...


    \displaystyle \lim_{x \to 0}\frac{\sin{(5x)} - 3x}{3x} = \lim_{x \to 0}\frac{\frac{d}{dx}[\sin{(5x)} - 3x]}{\frac{d}{dx}(3x)}

    \displaystyle = \lim_{x \to 0}\frac{5\cos{(5x)} - 3}{3}

    \displaystyle = \frac{5 - 3}{3}

    \displaystyle = \frac{2}{3}.
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  4. #4
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    Yes, but "overkill"! Like killing a mosquito with a 12 guage shot gun. I much prefer Liverpool's method.
    I also like the fact that Liverpool left something to be done rather than giving the answer- but I'll thank both of you since both replies were useful.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Yes, but "overkill"! Like killing a mosquito with a 12 guage shot gun. I much prefer Liverpool's method.
    I also like the fact that Liverpool left something to be done rather than giving the answer- but I'll thank both of you since both replies were useful.
    I prefer the term "brute force", and I consider it to be more effective than beating around the bush. It's like punching someone in the face or talking behind their back...
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  6. #6
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    But talking behind their back is much safer than punching them in the face!
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  7. #7
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    What would Chuck Norris do, I wonder ?

    Well, to answer the questions....

    If you want to split it up

    \displaystyle\frac{Sin5x-3x}{3x}=\frac{Sin5x}{3x}-\frac{3x}{3x}=\frac{Sin5x}{3x}-1

    So, to evaluate this limit, you only need evaluate \displaystyle\lim_{x \to 0}\frac{Sin5x}{3x}

    and then subtract 1.

    An alternative method is offered by Prove It "if" you are familiar with L'Hopital's Rule.
    Last edited by Archie Meade; January 22nd 2011 at 06:52 AM.
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