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Owlcoffee

  • one year ago

(1) Prove the limit of the sum of two function is equal to the sum of their independant limits. \[\lim_{x \rightarrow a} f(x)=M\] \[\lim_{x \rightarrow a} g(x)=L\] \[\lim_{x \rightarrow a}f(x) + g(x)=L +M\]

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  1. nincompoop
    • one year ago
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    |dw:1437984297705:dw|

  2. nincompoop
    • one year ago
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    |dw:1437984682123:dw|

  3. nincompoop
    • one year ago
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    |dw:1437984849692:dw|

  4. Owlcoffee
    • one year ago
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    Well, That is true but... I am supposed to prove the sum of limits... Not apply it.

  5. nincompoop
    • one year ago
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    using the limit definition is a proof too

  6. Owlcoffee
    • one year ago
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    You used the definition of derivative. I was thinking about using: \[f/a E(b, \epsilon ) \exists E ^{*}(a, \delta )/ \forall x \in E^* (a, \delta ) => f(x)\in E(b, \epsilon)\]

  7. UsukiDoll
    • one year ago
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    what is beta, delta, epsilon ?

  8. Owlcoffee
    • one year ago
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    The radius of the enviroments with center "b" (epsilon), and the one with center a (delta). I have thought that I need to prove that: \[\forall x \in E^* (a, \delta ): \left| f(x)+g(x)-(b +c) \right|< \epsilon \] It should suffice since it would mean that the definition applies.

  9. zzr0ck3r
    • one year ago
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    Let \(\epsilon >0 \) be given. \( \lim_{x \rightarrow a} f(x)=M\) implies \(\exists \delta_0 >0\) s.t. \(x\in (-\delta_0+a, \delta_0+a)\implies |f(x)-M| < \frac{\epsilon}{2}\). \(\lim_{x \rightarrow a} g(x)=L\) implies \(\exists \delta_1 >0\) s.t. \(x\in (-\delta_1+a, \delta_1+a)\implies |g(x)-L| < \frac{\epsilon}{2}\). Consider \(\delta = \min \{\delta_0, \delta_1\}\) and suppose \(|x-a|< \delta\), then obviously \(|x-a|< \delta_0\) and \(|x-a|<\delta_1\) so that \(|f(x)-M|<\frac{\epsilon}{2}\) and \(|g(x)-L|< \frac{\epsilon}{2}\). It follows from the triangle inequality that \[|f(x)+g(x)-(M+L)|=|f(x)-M+g(x)-L|\le |f(x) - M| + |g(x)-L| \\ < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\] as desired

  10. zzr0ck3r
    • one year ago
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    Let me know if you have any questions.

  11. Owlcoffee
    • one year ago
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    Haha, I had just solved it when you posted that, and I confirmed that my answer was correct, thanks.

  12. zzr0ck3r
    • one year ago
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    it follows from sequential definition of limits if you know it is true for sequences. So that would be a much shorter proof...

  13. nincompoop
    • one year ago
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    ya I learned that but I didn't really get into it too much because it was way over my brain to remember.

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