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Math Help - I still need help on 2 calculus problems

  1. #1
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    Exclamation I still need help on 2 calculus problems

    I have to use L'Hopital rule to figure out

    lim (1+2/x)^x
    x--)0

    Find Partial Fraction for

    -x^2-7x+27
    __________
    x(x^2+9)
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  2. #2
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    earboth's Avatar
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    Hi, nimmy,

    I cann't offer you a solution using the rule of de l'H˘pital, but maybe you can make some use of the following way:

    In

    \lim_\subc{x \rightarrow 0}{\left( 1+\frac{2}{x} \right)^x}

    you can substitute x=\frac{1}{n} because \lim_{\subc{n \rightarrow \infty}}{\left(\frac{1}{n} \right) = 0}

    Your limit changes to:

    \lim_{\subc{n \rightarrow \infty}}{\left( 1+2n \right)^{\frac{1}{n}} = 1

    For your 2nd question I've a question too: Are you sure that you didn't make a mistake in typing the term?

    Bye
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  3. #3
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    Do you know how to do partial fraction decomposition? Not to offend, but that shouldn't really be that difficult, unless I read it wrong.

    Set that whole thing equal to A/x + (Bx + C)/(x^2 + 9). Then multiply both sides by x*(x^2 + 9). You should then have
    -x^2 - 7x +27 = A(x^2 + 9) + Bx^2 + Cx.

    Don't ask for a proof, but I'm sure you can understand that the elements above must be grouped together according to the exponential power of x. The x^2 must be grouped together,etc. So.....

    -x^2 = Ax^2 + Bx^2
    -1 = A + B

    -7x = Cx
    C = -7

    27= 9A
    A = 3

    Little bit of the ole algebra and yer home free!
    That is how you do that, but there are other situations, like having (x-3)^2 in the denominator. Read though the strategy and I'm sure you'll get it.
    Last edited by MechHead; January 19th 2006 at 11:45 PM. Reason: Didn't look like I wanted it. Too confusing
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