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Math Help - continuity at a point

  1. #1
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    continuity at a point

    using the definition:  \forall \epsilon >0, \exists \delta >0, \forall h, \ |h|<\delta \ : |f(a+h)-f(a)|<\epsilon.

    Prove f(x)= x^{1/3}+x is continuous at the point a=0.


    given epsilon>0, pick delta , then given h with |h|<delta.

     \delta = \mbox{min}\{1, \epsilon^3/8\}

    we have |f(a+h)-f(a)|=  |h^{1/3} + h| \leq |h^{1/3}|+|h|<?=\epsilon

    does this look good? what shall write for ?
    Last edited by charikaar; December 3rd 2009 at 09:48 AM.
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  2. #2
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    Quote Originally Posted by charikaar View Post
    using the definition:  \forall \epsilon >0, \exists \delta >0, \forall h, \ |h|<\delta \ : |f(a+h)-f(a)|<\epsilon.

    Prove f(x)= x^{1/3}+x is continuous at the point a=0.


    given epsilon>0, pick delta , then given h with |h|<delta.

     \delta = \mbox{min}\{1, \epsilon^3/8\}

    we have |f(a+h)-f(a)|=  |h^{1/3} + h| \leq |h^{1/3}|+|h|<?=\epsilon

    does this look good? what shall write for ?
    If |h|<1 then what is the inequality between h^\frac{1}{3} and h?
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  3. #3
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    Thank you for your reply.

    how do we know this |h|<1

    I have  |h^{1/3} + h|^3 \leq 8|h| but i don't understand why?
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  4. #4
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    You can restrict your delta. If you let  \delta < 1 then it follows that  |h| < \delta < 1 .

    Remember that  2 \delta is the length of your interval, and so by making sure  \delta < 1 you're restricting the interval so its length is less than 2.

    Remember, you only care about how the function behaves in some neighborhood about your point, so this restriction is okay.
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  5. #5
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    Quote Originally Posted by charikaar View Post
    using the definition:  \forall \epsilon >0, \exists \delta >0, \forall h, \ |h|<\delta \ : |f(a+h)-f(a)|<\epsilon.

    Prove f(x)= x^{1/3}+x is continuous at the point a=0.
    given epsilon>0, pick delta , then given h with |h|<delta.
     \delta = \mbox{min}\{1, \epsilon^3/8\}
    we have |f(a+h)-f(a)|=  |h^{1/3} + h| \leq |h^{1/3}|+|h|<?=\epsilon
    does this look good? what shall write for ?
    You have some basic mistakes.
    Recall that a=0 so that |f(a+h)-f(a)|=|f(0+h)-f(0)|=|h^{\frac{1}{3}}|<\delta^{\frac{1}{3}}.
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  6. #6
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    Just for closure I will mention.

    Quote Originally Posted by charikaar View Post
    how do we know this |h|<1
    If you look at how you picked delta, it guarantees that h will be less than 1.
    Quote Originally Posted by charikaar View Post
    given epsilon>0, pick delta , then given h with |h|<delta.

     \delta = \mbox{min}\{1, \epsilon^3/8\}
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  7. #7
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    Quote Originally Posted by Plato View Post
    You have some basic mistakes.
    Recall that a=0 so that |f(a+h)-f(a)|=|f(0+h)-f(0)|=|h^{\frac{1}{3}}|<\delta^{\frac{1}{3}}.
    how did u get that? shouldnt
    |f(h)|=|h^{\frac{1}{3}}+h|
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  8. #8
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    ravenboy you are right!

    did you solve the question further?
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  9. #9
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    Quote Originally Posted by ravenboy View Post
    how did u get that? shouldnt
    |f(h)|=|h^{\frac{1}{3}}+h|
    Yes you are correct.
    In that case just pick  \delta = \mbox{min}\{1, \epsilon^3/8, \epsilon/2\}
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