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Math Help - Prove limit of x^5 = a^5 using epsilon delta

  1. #1
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    Prove limit of x^5 = a^5 using epsilon delta

    \lim_{x\to a}x^5=a^5

    I understand that the power rule for limits could be used here but my teacher wants it done using the epsilon delta definition.

    1st attempt (though I think it's obviously wrong since my \delta could be undefined):

    |x-a|< \delta \rightarrow              |x^5-a^5|< \epsilon
    |x^5-a^5|=|(x-a)(x^4+ax^3+a^2x^2+a^3x^3 +a^4)|< \epsilon
    |x-a|< \frac{\epsilon}{|(x^4+ax^3+a^2x^2+a^3x^3 +a^4)|}

    Choose:    \delta = \frac{\epsilon}{|(x^4+ax^3+a^2x^2+a^3x^3 +a^4)|}
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  2. #2
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    Re: Prove limit of x^5 = a^5 using epsilon delta

    You're right that you need to factor. But, you shouldn't have x terms in your choice of delta. Try getting a bound on the big factor (x^4+...) by making a bound on x.
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  3. #3
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    Re: Prove limit of x^5 = a^5 using epsilon delta

    Do you mean something very similar to the below example?
    Are you suggesting the bound I make is something like |x-a|<1?
    which implies -1<x-a<1

    a-1<x<a+1

    x^4<(a+1)^4

    |x^3|<|(a+1)^3|


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    Re: Prove limit of x^5 = a^5 using epsilon delta

    Indeed. But don't forget that you want an upper bound, and you certainly have that x< |a|+1 So, use this to get aqn upper bound on the big polynomial factor, take delta to be the minimum of 1 and epsilon/(bound on poly)
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