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Math Help - calculus help on Limits and epsilon delta

  1. #1
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    calculus help on Limits and epsilon delta

    Hi, I do need your help with the question attached above.

    I do have another question:

    Prove by the eplison delta method

    Lim of X^4 when x approaches -2 equal to 16
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by guess View Post
    Hi, I do need your help with the question attached above.

    I do have another question:

    Prove by the eplison delta method

    Lim of X^4 when x approaches -2 equal to 16
    I'll do the \varepsilon-\delta problem.

    \lim_{x\to-2}x^4=16

    The objective is to find \delta such that 0<\left|x^4-16\right|<\delta whenever \left|x+2\right|<\varepsilon.

    Note that \left|x^4-16\right|=\left|(x^2+4)(x^2-4)\right|=\left|(x^2+4)(x-2)(x+2)\right|<\delta.

    Let us supposed \left|x+2\right|<1. We find the upper bounds on \left|x-2\right| and \left|x^2+4\right|

    \left|x-2\right|=\left|x+2-4\right|\leq\left|x+2\right|+\left|-4\right|=1+4=5

    \left|x^2+4\right|=\left|x^2-4+8\right|\leq\left|x-2\right|\left|x+2\right|+\left|8\right|=(5)(1)+8=1  3.

    Thus, \left|x^4-16\right|<13\left|x+2\right|<\varepsilon\implies \left|x+2\right|<\frac{\varepsilon}{13}.

    If we take \delta=\frac{\varepsilon}{13}, the proof is complete.

    Does this make sense?
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  3. #3
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    hi,

    correct me if i'm wrong:

    |X^4-16|= |(X+2)(X-2)(X^2+4)|

    |X-2|=|X+2-4|<|X+2|+|-4|

    we let delta less then 1

    |X+2|< 1

    |X-2| < (3)+(4) = 7

    |X^2-4+8|<|(X+2)(X-2)+8|<|X+2||X-2|+|8| = (1)(5)+8 =13

    from here,

    |X^4-16|= |(X+2)(X-2)(X^2+4)|< delta(7)(13) = 91 delta

    91 delta = epsilon
    delta = epsilon/91.

    We chose the min delta {1, epsilon/91}
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post

    If we take \delta=\min\left\{1,\frac{\varepsilon}{13}\right\}, the proof is complete
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