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Math Help - Finding the Values Using a Limit

  1. #1
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    Finding the Values Using a Limit

    Hi,

    I'm not sure how to find out the answer to this qustion so any help would be great. I have a hunch that K may be -1 and l may be +/-4.

    Thanks, DLL.
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  2. #2
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    Try replacing \cos lx with the Maclaurin series for it.
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  3. #3
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    Hi,

    I tried doing the Maclaurin series but I'm not sure where to go from there. I can't see where finding the values a k and l comes into it.

    Thanks, DLL
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  4. #4
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    Quote Originally Posted by Daddy_Long_Legs View Post
    Hi,

    I'm not sure how to find out the answer to this qustion so any help would be great. I have a hunch that K may be -1 and l may be +/-4.

    Thanks, DLL.
    The denominator is going to 0 so, in order to have any finite limit, the numerator must also go to 0. Since cos(x) goes to 1 as x goes to 0, yes, k must be -1.

    You then have \frac{-1+ cos(lx)}{x^2}. I would suggest using L'Hopital's rule: differentiating numerator and denominator separately, we get \frac{-l sin(lx)}{2x}= -\frac{l^2}{2}\frac{sin(lx)}{lx} where I have multiplied numerator and denominator by l to get "sin(z)/z" as z goes to 0.
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  5. #5
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    Maclaurin series: \cos lx=1-\frac{(lx)^2}{2}+O(x^4)

    Put that into the limit and you get \lim_{x \to 0}\frac{k+1-\frac{(lx)^2}{2}+O(x^4)}{x^2}=\frac{k+1}{x^2}-\frac{\frac{(lx)^2}{2}+O(x^4)}{x^2}=-4

    We can see that k=-1 since otherwise the limit wouldn't be finite.

    With that out of the way, we have \lim_{x \to 0}-\frac{\frac{(lx)^2}{2}+O(x^4)}{x^2}=\lim_{x \to 0}\left(-\frac{l^2}{2}+O(x^2)\right)=-4 \Longleftrightarrow l^2=8 since \lim_{x \to 0} O(x^2)=0
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  6. #6
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    HallsofIvy

    I tried doing the problem the way you sugested but I'm not sure how to find the value of l after you get to "sin(z)/z" as z goes to 0. I reasonably certain the value of l is -/+4. I just can't seem to find how to find it.

    Thanks.
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  7. #7
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    Quote Originally Posted by Daddy_Long_Legs View Post
    HallsofIvy

    I tried doing the problem the way you sugested but I'm not sure how to find the value of l after you get to "sin(z)/z" as z goes to 0. I reasonably certain the value of l is -/+4. I just can't seem to find how to find it.

    Thanks.
    If you're going to use l'Hopital's rule once, you may as well use it twice.

    Using it the second time gives \lim_{x \rightarrow 0} \frac{-l^2 \cos (lx)}{2}. So \frac{-l^2}{2} = -4 ....


    By the way, you're expected to know that \lim_{x \rightarrow 0} \frac{\sin z}{z} = 1 ....
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