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Math Help - Limits, absolute values, and inequalities with variables in the denominator

  1. #1
    Member Chokfull's Avatar
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    Limits, absolute values, and inequalities with variables in the denominator

    Here is the original problem
    <br /> <br />
\lim_{x\to-1^-}\frac {5} {(x+1)^3}= \infty<br />

    So you do
    \frac {5} {(x+3)}<N whenever -1-\delta <x<-1

    Then you need to solve for x+1 to move on to the next step, but I don't know how...
    The book says you should get \sqrt [3] {5/|N|}... but I can't figure out how to get the absolute value.
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  2. #2
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    Better would be to write \frac 5{(x+1)^3}<-N ( N positive) when -1-\delta<x<-1 because \lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty.
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  3. #3
    Member Chokfull's Avatar
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    Quote Originally Posted by halbard View Post
    Better would be to write \frac 5{(x+1)^3}<-N ( N positive) when -1-\delta<x<-1 because \lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty.
    OK... but that still doesn't answer my question
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  4. #4
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    Quote Originally Posted by Chokfull View Post
    Here is the original problem
    [FONT=monospace] <br /> <br />
\lim_{x\to-1^-}\frac {5} {(x+1)^3}= \infty<br />
    The orginal problem is wrong.
    \lim_{x\to-1^-}\frac {5} {(x+1)^3}= {\color{red}-}\infty<br />

    Suppose that N \in \mathbb{Z}^ +  then \sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x <  - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} <  - N.
    Last edited by Plato; July 13th 2009 at 03:22 PM.
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  5. #5
    Member Chokfull's Avatar
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    Quote Originally Posted by Plato View Post
    The orginal problem is wrong.
    \lim_{x\to-1^-}\frac {5} {(x+1)^3}= {\color{red}-}\infty<br />

    Suppose that N \in \mathbb{Z}^ +  then \sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x <  - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} <  - N.
    I think that the problem may have been to prove that this limit is false... that or it was a typo by me or the book... or you are mistaken.
    Anyways, does that Z mean positive? if it does, you are wrong--N has to be negative.
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  6. #6
    Member Chokfull's Avatar
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    I'm trying to use the book's method to solve this problem.... The part I need help with is the absolute value in the denominator, I thought there was a rule for inequalities in which you can't multiply each side by the denominator, but I couldn't find it.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    \lim_{x\to{a}}f(x)=\infty\Rightarrow that for each number \delta such that |x-a|<\delta, \exists a number M for which M<f(x).

    You can see that the definition is satisfied.However, f(x) goes to NEGATIVE infinity. If you would like to prove it you could choose some arbitraily small number M, solve the inequality M>f(x) , then use that value of x (to the left of -1) in the inequality |x-a|<\delta=\text{slightly larger than the actual value} (where a is -1), and show that however close x comes to a, f(x) will still be less than M.
    Last edited by VonNemo19; July 16th 2009 at 01:43 PM.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Plato View Post
    The orginal problem is wrong.
    x\rightarrow{-1^-},\forall{x}\epsilon(-\infty,-1),\Rightarrow(x+1)^3<0\Rightarrow{f(x)<0}

    You are correct sir.
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  9. #9
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    \frac{5}{(x+1)^{3}}\rightarrow<br />
-\infty as x\rightarrow -1^{-} if |x+1|<\delta whenever \frac{5}{(x+1)^{3}}<-N, N>0 which yields
    after solving the absolute value that \frac{5}{(x+1)^{3}}<\frac{-5}{\delta^{3}} so we can write \frac{-5}{\delta^{3}}=-N which directly implies the result.
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