I can't understand this limit of a sum proof I'm reading a text which tries to prove that $$\lim_{x\to a} (f+g)=\lim_{x\to a}f+\lim_{x\to a}g$$
the first step goes like this "Let $\epsilon >0$, then because $\lim_{x\to a}f=k$ and $ \lim_{x\to a}g=L $ there is a $\delta_1>0$ and $\delta_2>0$, such that $$\vert f - K\vert<\frac{\epsilon}{2}$$
$$\vert g - L\vert<\frac{\epsilon}{2}$$
whenever $0<\vert x-a\vert \ <\delta_1$ and $0<\vert x-a\vert \ <\delta_2$"
Why is the distance to the limit not less than $\epsilon$? Why is it written as $\frac{\epsilon}{2}$, what is the rationale behind this?
 A: This is the rationale.
We arbitrarily pick an $\epsilon > 0$.
We want to find a $\delta$ where where $|x-a|< \delta$ will mean that $|(f(x)+g(x)) -(K+L)| < \epsilon$.  how will we do that?
Well, $|(f(x)+g(x)) -(K+L)|=|(f(x)-K)+(g(x)-L)| \le |f(x)-K| + |g(x)-L|$.  SO if we can prove that 
$|f(x)-K| + |g(x)-L| < \epsilon$ we will be done.
Now $f(x)\to K$ so for any $\overline \epsilon  > 0$ we can find a $\delta_1$ so that $|x-a|< \delta_1$ implies $|f(x)-K| < \overline \epsilon$.
And $g(x)\to L$ so for any $\epsilon' > 0$ we can find a $\delta_2$ so that $|x-a| < \delta_2$ implies $|g(x)-L|< \epsilon'$.
So if $|x-a| < \min(\delta_1, \delta_2)$ the we would have
$|f(x)-K| + |g(x)-L| < \overline \epsilon + \epsilon'$.
But we WANT $|f(x)-K| + |g(x)-L| < \epsilon$.
So if we can have $\overline \epsilon + \epsilon' = \epsilon$ we will be done.
... that is our proof would be.....

For $\epsilon > 0$ let $\overline \epsilon$ be so that $0 < \overline \epsilon < \epsilon$ and let $\epsilon' = \epsilon -\overline \epsilon$ so that $\epsilon = \overline \epsilon + \epsilon'$
There exist $\delta_1$ and $\delta_2$ so that if $|x-a| < \delta_1$ then $|f(x)-K|< \overline \epsilon$ and if $|x-a| < \delta_2$ then $|g(x) -L|< \epsilon'$.
So if $\delta = \min(\delta_1, \delta_2$ then $|x-a|< \delta$ means $|x-a|< \delta_1$ so $|f(x)-K|< \overline \epsilon$ and as well, $|x-a|< \delta_2$ so $|g(x)-L|< \epsilon'$.
And so $|x-a|<\delta$ means $|(f+g)(x) - (K+L)| \le |f(x)-K|+|g(x)-L| < \overline \epsilon + \epsilon' = \epsilon$.
So $\lim_{x\to a} (f+g)(x) = K + L$.

Now the only thing left is to figure out for any arbitrary positive $\epsilon$ how do we find positive $\overline \epsilon, \epsilon'$ so that $\overline\epsilon + \epsilon' = \epsilon$?
And... well,.... if $\overline \epsilon = \frac {\epsilon}2$ and $\epsilon' = \frac {\epsilon}2$ then $\overline \epsilon + \epsilon' =\epsilon$ is a good choice.  We could have chosen any other pair though.
A: The goal is to get the final inequality to be of the form "${\dots} < \varepsilon$".  This will occur when the pair of $\frac{\varepsilon}{2}$s that you ask about are added (which will most likely occur in an application of the triangle inequality).
