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Thread: 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
    $\displaystyle

    $
    $\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= \infty
    $

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

    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 $\displaystyle \sqrt [3] {5/|N|}$... but I can't figure out how to get the absolute value.
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    Better would be to write $\displaystyle \frac 5{(x+1)^3}<-N$ ($\displaystyle N$ positive) when $\displaystyle -1-\delta<x<-1$ because $\displaystyle \lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty$.
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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 $\displaystyle \frac 5{(x+1)^3}<-N$ ($\displaystyle N$ positive) when $\displaystyle -1-\delta<x<-1$ because $\displaystyle \lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty$.
    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]$\displaystyle

    $
    $\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= \infty
    $
    The orginal problem is wrong.
    $\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= {\color{red}-}\infty
    $

    Suppose that $\displaystyle N \in \mathbb{Z}^ + $ then $\displaystyle \sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x < - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} < - N$.
    Last edited by Plato; Jul 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.
    $\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= {\color{red}-}\infty
    $

    Suppose that $\displaystyle N \in \mathbb{Z}^ + $ then $\displaystyle \sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x < - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} < - N$.
    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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    $\displaystyle \lim_{x\to{a}}f(x)=\infty\Rightarrow$ that for each number $\displaystyle \delta$ such that $\displaystyle |x-a|<\delta$, $\displaystyle \exists$ a number $\displaystyle M$ for which $\displaystyle M<f(x)$.

    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 $\displaystyle M$, solve the inequality $\displaystyle M>f(x)$ , then use that value of x (to the left of -1) in the inequality $\displaystyle |x-a|<\delta=\text{slightly larger than the actual value}$ (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; Jul 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.
    $\displaystyle x\rightarrow{-1^-},\forall{x}\epsilon(-\infty,-1),\Rightarrow(x+1)^3<0\Rightarrow{f(x)<0}$

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