About limit theorems So I decided to go back and study the proofs about limit theorems. This one stumped me;

Theorem:  If $\displaystyle\lim_{x\to a} f(x)=L$ and $\displaystyle\lim_{x\to a} g(x)=M$ then $\displaystyle\lim_{x\to a}[f(x)+g(x)]=L+M$.

I wonder if I can link? Anyway, the proof is here: http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx
at the 'proof of 2' note. My question is, why do we have to choose the $\delta$ as the smaller of the two; $\delta _1$ and $\delta_2$? Also, why is it =$\epsilon $ at last part, shouldn't it be 
 A: The reason we choose the smaller $\delta$ is because only the smaller one "covers" both cases (for both $f$ and $g$).  
Let me illustrate by example.  Let $f(x) = 2x$ and $g(x) = 3x$, and consider the (true) statements,
$$ \lim_{x \to 0} 2x = 0, \qquad \lim_{x \to 0} 3x = 0.$$
We wish to prove that
$$ \lim_{x \to 0} (2x + 3x) = \lim_{x\to 0} 2x + \lim_{x \to 0} 3x 
= 0 + 0 = 0.$$
In the proof (referenced by the link you provided), the first step is to pick arbitrary $\epsilon > 0$ and find the corresponding $\delta_1, \delta_2$ that cause the value of each function to be within $\epsilon/2$ of each limit value.  That is:
$$ |f(x) - L| < \epsilon/2, \;\Rightarrow\; |2x| < \epsilon/2
\;\Rightarrow\; |x| < \epsilon/4.$$
Thus we choose $\delta_1 = \epsilon/4$.
$$ |g(x) - M| < \epsilon/2, \;\Rightarrow\; |3x| < \epsilon/2
\;\Rightarrow\; |x| < \epsilon/6.$$
Thus, $\delta_2 = \epsilon/6$.  Now consider if we chose the larger instead of smaller $\delta_i$.  Let $\delta = \epsilon/4$.  The "proof" of the limit $\lim_{x\to 0} (f(x) + g(x)) = 0$ would go something like this:
Let $\epsilon > 0$ be arbitrary.  Fix $\delta = \epsilon/4$, and let $x$ be such that $|x| < \delta$.
$$ |f(x) + g(x) - (L+M)| = |5x - 0| = 5|x| < 5\delta = \frac{5}{4}\epsilon.$$
Oops!!  We wanted to show the expression is less than $\epsilon$, and unfortunately this "proof" does not provide air-tight evidence for the limit.  However, if we used the smaller $\delta = \epsilon/6$, it would have worked out fine.
In general, of course, we do not have any control over what $f$ and $g$ are, so we make the most restrictive choice for $\delta$ in order to make an air-tight case for the limit of the sum.
Hope this helps!
A: We choose the smallest $\delta$ so that the inequality is as tight as possible. (Basically, Shawn explained that part well.) Now, to explain the other Q:
This is primarily coming from how concise Paul's proof style is (in that particular part). He's saying:
$$\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
However, the way he wrote it, it's ambiguous. What I think makes more sense is:
\begin{align}
|f(x)+g(x)-(K+L)|&=|(f(x)-K)+(g(x)-L)| \quad \text{just rewriting}\\
|(f(x)-K)+(g(x)-L)|&\leq |(f(x)-K)|+|(g(x)-L)| \quad \text{via } |a+b|\leq |a|+|b|\\
|(f(x)-K)|+|(g(x)-L)|&<\frac{\epsilon}{2}+\frac{\epsilon}{2} \quad \text{See below}\\
\frac{\epsilon}{2}+\frac{\epsilon}{2}&=\epsilon\\
\therefore |(f(x)-K)|+|(g(x)-L)|&<\epsilon\\
&\text{Q.E.D.}
\end{align}
Line $3$ follows from adding the two inequalities together:
$$|(f(x)-K|<\frac{\epsilon}{2} \text{ and } |g(x)-L|<\frac{\epsilon}{2}$$
