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Math Help - Limits Problems

  1. #1
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    Limits Problems

    Hi. I need help. My test is next week, and on it there will be a lot of math stuff including limits. Now, the teacher gave us a few problems to show us what to expect on the test. That was last week. I've been trying to solve these limits:

    Limits Problems-problem-2.bmpLimits Problems-problem-3.bmpLimits Problems-problem-1.bmp



    but without success.

    Here's the catch: I can't use l'Hopital's Rule.

    I would really appreciate some solutions. Thx in advance.

    P.S. You don't have to write the whole solutions, just the general idea.
    Last edited by StanceGlanceAdvance; April 5th 2011 at 01:01 AM.
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  2. #2
    Senior Member Sambit's Avatar
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    For the "sin" problem, multiply numerator and denominator by \sqrt{y}+1 and note that \lim_{y\to0}\frac{\sin y}{y}=1, the answer will be -2

    For the "ln" problem, note that x approaches more rapidly to 0 than \ln x, so the answer becomes 0.

    The other one--algebraic manipulation.
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  3. #3
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    Hello, StanceGlanceAdvance!

    \displaystyle \lim_{x\to4} \frac{x-4}{\sqrt[3]{x+23} - 3}

    First, recall the identity: . a^3 - b^3 \:=\:(a-b)(a^2+ab + b^2)


    We have: . \displaystyle f(x) \;=\;\frac{x-4}{(x+23)^{\frac{1}{3}} - 3}


    Multiply numerator and denominator by: (x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9

    \displaystyle f(x) \;=\;\frac{x-4}{(x+23)^{\frac{1}{3}} - 3}\cdot\frac{(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9} {(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9}

    . . . . \displaystyle =\;\frac{(x-4)\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]}{(x+23) - 3^3}

    . . . . \displaystyle =\;\frac{(x-4)\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]}{x-4}

    . . . . \displaystyle =\;(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9


    Therefore: . \displaystyle \lim_{x\to4}\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]

    . . . . . . . . . . =\; 27^{\frac{2}{3}} + 3\cdot27^{\frac{1}{3}} + 9

    . . . . . . . . . . =\;9 + 3\cdot3 + 9

    . . . . . . . . . . = \;27

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  4. #4
    MHF Contributor chisigma's Avatar
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    Setting x=1-\xi the limit becomes...

    \displaystyle \lim_{x \rightarrow 0} x\ \ln x = \lim_{\xi \rightarrow 1} (1-\xi)\ \ln (1-\xi) (1)

    Now is...

    \displaystyle \ln (1-\xi) = -\xi - \frac{\xi^{2}}{2} - \frac{\xi^{3}}{3} - ... (2)

    ... so that...

    \displaystyle (1-\xi)\ \ln (1-\xi) = -\xi\ (1-\xi) - \frac{\xi^{2}}{2}\ (1-\xi) - \frac{\xi^{3}}{3}\ (1-\xi) - ... (3)

    If \xi = 1 each term of the 'infinite sum' (3) is 0 so that the limit of 'infinite sum' is 0...

    Kind regards

    \chi \sigma
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  5. #5
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    You guys are amazing... I love this forum.

    Thank you so much. Thanks allround!
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