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I was reading the following proof about $\lim_{x\to a}$$[f(x)+g(x)]$$=$$\lim_{x\to a}f(x)$$+$$\lim_{x\to a}g(x)$

In the proof it is assumed that $\lim_{x\to a}f(x)=K$ and $\lim_{x\to a}g(x)=L$

Here is the proof:

Let $\epsilon$>0 then because $\lim_{x\to a}f(x)=K$ and $\lim_{x\to a}g(x)=L$ there is an $\delta_1$ and $\delta_2$ such that,

$\vert f(x)-K\vert$<$\frac \epsilon2$ whenever $0<\vert x-a\vert<\delta_1$ and $\vert g(x)-L\vert$<$\frac \epsilon2$ whenever $0<\vert x-a\vert<\delta_2$

Now choose $\delta= \min \{\delta_1,\delta_2\}$

We need to show that "$\vert f(x)+g(x)-(K+L)\vert < \epsilon$" whenever $0<\vert x-a\vert<\delta$

This is the part I don't understand.First how he got this absolute value from the left side? Can I just add the two absolute values from before to get this result? I know that if I add $\frac \epsilon2$+$\frac \epsilon2$ I get the right side of inequality but I don't know how to get the one from the left

Assume that $0<\vert x-a\vert<\delta$, we then have $\vert f(x)+g(x)-(K+L)\vert = \vert (f(x)-K)+(g(x)-L)\vert $$<=\vert f(x)-K \vert + \vert g(x)-L \vert$ (by triangle inequality) $<\frac \epsilon2+\frac \epsilon2$(by previous definition)

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    Your bold-letter question is odd: there's written *explicitly* that that is what's needed to be proved, not that "he got whatever" ! The point here is: do **you** understand why is that what's needed to be proved? – 2017-02-07
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    Please indicate in bold (or quotes, maybe) the mathematical statement you do not understand, specifically, because I still find your question unclear. The left side's absolute value comes from trying to prove by definition that something is a limit, and then the triangle inequality. I do not see where the problem is, so if you could highlight it it would be nice. – 2017-02-07
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    i just quoted the part i dont understand, I dont get how he got that absolute value part. Hope I am more clear know – 2017-02-07
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    @kprincipe Do you know that $\;\lim_{x\to x_o}f(x)=L\implies \;$ for all $\;\epsilon>0\;$ exists $\;\delta>0\;$ such that if $\;|x-x_0|<\delta\;$ then $\;|f(x)-L|<\epsilon\;$ ? Well, the proof of what you were reading (look at the beginning of your question!) says that $\;\lim(f(x)+g(x))=K+L\;$ , so for all...etc. .... **you must prove that** $\;|(f(x)-g(x))-(K+L)|<\epsilon\;$ .... – 2017-02-07

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