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Math Help - Limit confusion

  1. #1
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    Limit confusion

    This problem says if n>=3, derive the inequality -x^2 <= x^n <= x^2 for -1 < x < 1. Not really sure how to even start this...
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  2. #2
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    -1 \leq x \leq 1 multiplying both sides by x^2

    -x^2 \leq x^3 \leq x^2

    I think that you can prove the general result for n\geq 3 using induction
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  3. #3
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    Quote Originally Posted by felper View Post
    -1 \leq x \leq 1 multiplying both sides by x^2

    -x^2 \leq x^3 \leq x^2

    I think that you can prove the general result for n\geq 3 using induction
    Ok I'm kind of stuck, it's been a while since I've used induction. I'm at the point where -x^2 \leq x^{(n+1)} \leq x^2 . I'm thinking about dividing by through by x...
    Last edited by Plato; February 4th 2010 at 07:05 AM.
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  4. #4
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    We proved that the inequality is true for n=3. Now, suppose that is true for n

    -x^2\leq x^n\leq x^2 (hypothesis)

    Now

    -1\leq x \leq 1 multiply by x^n both sides

    -x^n \leq x^{n+1} \leq x^n

    Using the hypothesis

    -x^2\leq -x^n \leq x^{n+1} \leq x^n \leq x^2 \Rightarrow -x^2\leq x^{n+1}\leq x^2 \quad \Box
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  5. #5
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    Quote Originally Posted by felper View Post
    -x^2\leq -x^n \leq x^{n+1} \leq x^n \leq x^2 \Rightarrow -x^2\leq x^{n+1}\leq x^2 \quad \Box
    How do you know -x^2\leq -x^n
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  6. #6
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    Hay Cat here is what is going on.
    -1<x<1\text{ if and only if }|x|<1.
    Now recall this \left| x \right|\left| {x^n } \right| = \left| {x^{n + 1} } \right|.
    We can say the x\ne 0. For if it were there is nothing to prove.
    \left| x \right| < 1 \Rightarrow \left| {x^2 } \right| < \left| x \right| < 1 \Rightarrow \left| {x^3 } \right| < \left| {x^2 } \right| < \left| x \right| < 1
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  7. #7
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    Ah yes, thanks for the help.
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