## Owlcoffee one year ago (1) Prove the limit of the sum of two function is equal to the sum of their independant limits. $\lim_{x \rightarrow a} f(x)=M$ $\lim_{x \rightarrow a} g(x)=L$ $\lim_{x \rightarrow a}f(x) + g(x)=L +M$

1. nincompoop

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2. nincompoop

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3. nincompoop

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4. Owlcoffee

Well, That is true but... I am supposed to prove the sum of limits... Not apply it.

5. nincompoop

using the limit definition is a proof too

6. Owlcoffee

You used the definition of derivative. I was thinking about using: $f/a E(b, \epsilon ) \exists E ^{*}(a, \delta )/ \forall x \in E^* (a, \delta ) => f(x)\in E(b, \epsilon)$

7. UsukiDoll

what is beta, delta, epsilon ?

8. 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.

9. 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

10. zzr0ck3r

Let me know if you have any questions.

11. Owlcoffee

Haha, I had just solved it when you posted that, and I confirmed that my answer was correct, thanks.

12. zzr0ck3r

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

ya I learned that but I didn't really get into it too much because it was way over my brain to remember.