Owlcoffee
  • Owlcoffee
(1) Prove the limit of the sum of two function is equal to the sum of their independant limits. \[\lim_{x \rightarrow a} f(x)=M\] \[\lim_{x \rightarrow a} g(x)=L\] \[\lim_{x \rightarrow a}f(x) + g(x)=L +M\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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nincompoop
  • nincompoop
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nincompoop
  • nincompoop
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nincompoop
  • nincompoop
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Owlcoffee
  • Owlcoffee
Well, That is true but... I am supposed to prove the sum of limits... Not apply it.
nincompoop
  • nincompoop
using the limit definition is a proof too
Owlcoffee
  • Owlcoffee
You used the definition of derivative. I was thinking about using: \[f/a E(b, \epsilon ) \exists E ^{*}(a, \delta )/ \forall x \in E^* (a, \delta ) => f(x)\in E(b, \epsilon)\]
UsukiDoll
  • UsukiDoll
what is beta, delta, epsilon ?
Owlcoffee
  • Owlcoffee
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.
zzr0ck3r
  • zzr0ck3r
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
zzr0ck3r
  • zzr0ck3r
Let me know if you have any questions.
Owlcoffee
  • Owlcoffee
Haha, I had just solved it when you posted that, and I confirmed that my answer was correct, thanks.
zzr0ck3r
  • zzr0ck3r
it follows from sequential definition of limits if you know it is true for sequences. So that would be a much shorter proof...
nincompoop
  • nincompoop
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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