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Math Help - Finding Delta

  1. #1
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    Finding Delta

    Use the precise definition to prove the following limit. Given epsilon, find a suitable delta to prove
    lim 5x^2-3x-2 = 34
    x->-3

    So I got
    5x^2-3x-36
    |x+3| * |5x-12|

    -1 < x+3 < 1
    -4 < x < -2
    -20 < 5x < -10
    -32 < 5x-12 < -22
    How do I find delta if my delta is negative?

    Need help with these two as well.

    Tried multiplying the conjugate with this one but ran into a dead end.
    lim (Sqrt(2+x^2) - Sqrt(2-x^2)) / x^2
    x -> 0

    lim (x^(2/3)-4) / (x^(1/3)-2)
    x -> 8



    Thanks for the help!
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  2. #2
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    Quote Originally Posted by beanus View Post
    Use the precise definition to prove the following limit. Given epsilon, find a suitable delta to prove
    lim 5x^2-3x-2 = 34
    x->-3

    So I got
    5x^2-3x-36
    |x+3| * |5x-12|

    -1 < x+3 < 1
    -4 < x < -2
    -20 < 5x < -10
    -32 < 5x-12 < -22
    How do I find delta if my delta is negative?

    Need help with these two as well.

    Tried multiplying the conjugate with this one but ran into a dead end.
    lim (Sqrt(2+x^2) - Sqrt(2-x^2)) / x^2
    x -> 0

    lim (x^(2/3)-4) / (x^(1/3)-2)
    x -> 8



    Thanks for the help!
    Never mind you did. I couldn't tell at first.

    Here are some worked out examples

    http://www.math.ucdavis.edu/~kouba/C...l#SOLUTION%202
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  3. #3
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    First of all, \displaystyle \lim_{x \to -3}(5x^2 - 3x - 2) \neq 34, it equals \displaystyle 52.

    However, \displaystyle \lim_{x \to 3}(5x^2 - 3x - 2) = 34.


    So, to prove \displaystyle \lim_{x \to 3}(5x^2 - 3x - 2) = 34, you need to show that \displaystyle |x - 3| < \delta \implies |5x^2 - 3x - 2 - 34| < \epsilon for all \displaystyle \epsilon > 0.

    So solve \displaystyle |5x^2 - 3x - 2 - 34| < \epsilon for \displaystyle |x - 3|.


    \displaystyle |5x^2 - 3x - 36| < \epsilon

    \displaystyle |x-3||5x+12| < \epsilon.

    \displaystyle |x - 3| < \frac{\epsilon}{5x + 12}.


    Now, since taking a limit requires us to close in on the point \displaystyle x = 3, we can define \displaystyle \delta to be a small value, e.g. \displaystyle 1, since we are going to make it smaller anyway...

    So by defining \displaystyle \delta = 1, we have

    \displaystyle |x - 3| < 1

    \displaystyle -1 < x-3 < 1

    \displaystyle -5 < 5x - 15 < 5

    \displaystyle 10 < 5x < 20

    \displaystyle 22 < 5x + 12 < 32


    So we can say \displaystyle |x - 3| < \frac{\epsilon}{32}, provided \displaystyle |x - 3| < 1.


    So define \displaystyle \delta = \min\left\{1, \frac{\epsilon}{32}\right\}, and the proof will follow.
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  4. #4
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    Quote Originally Posted by beanus View Post
    Need help with these two as well.

    Tried multiplying the conjugate with this one but ran into a dead end.
    lim (Sqrt(2+x^2) - Sqrt(2-x^2)) / x^2
    x -> 0

    lim (x^(2/3)-4) / (x^(1/3)-2)
    x -> 8
    For these are you expected to use the precise definition?
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  5. #5
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    They just say "Find:"
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  6. #6
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    \displaystyle \lim_{x \to 0}\frac{\sqrt{2+x^2} - \sqrt{2-x^2}}{x^2} = \lim_{x \to 0}\frac{(\sqrt{2+x^2}-\sqrt{2-x^2})(\sqrt{2+x^2}+\sqrt{2-x^2})}{x^2(\sqrt{2+x^2}+\sqrt{2-x^2})}

    \displaystyle = \lim_{x \to 0}\frac{(2+x^2)-(2-x^2)}{x^2(\sqrt{2+x^2}+\sqrt{2-x^2})}

    \displaystyle = \lim_{x \to 0}\frac{2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}.

    I'm sure you can go from here...
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  7. #7
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    Thanks! I get that one. Any thing on the last one? I multiplied by the conjugate and got

    (x^4/3 - 16) / ((x^1/3 - 2) (x^2/3 + 4))
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  8. #8
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    Multiplying by the conjugate won't help you with the second...

    Note that \displaystyle x^{\frac{2}{3}} - 4 = \left(x^{\frac{1}{3}}\right)^2 - 2^2 = \left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} + 2\right).

    Something cancels...
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