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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.
\[\lim_{x \rightarrow a} f(x)=M\]
\[\lim_{x \rightarrow a} g(x)=L\]
\[\lim_{x \rightarrow a}f(x) + g(x)=L +M\]
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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nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437984297705:dw

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437984682123:dw

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437984849692:dw

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0Well, That is true but... I am supposed to prove the sum of limits... Not apply it.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0using the limit definition is a proof too

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0what is beta, delta, epsilon ?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.2Let \(\epsilon >0 \) 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 \(xa< \delta\), then obviously \(xa< \delta_0\) and \(xa<\delta_1\) 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
 one year ago
Best ResponseYou've already chosen the best response.2Let me know if you have any questions.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0Haha, I had just solved it when you posted that, and I confirmed that my answer was correct, thanks.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2it follows from sequential definition of limits if you know it is true for sequences. So that would be a much shorter proof...

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0ya 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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