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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

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    @Dystopian: If you do what Salech noted, you will not go any further in the Theorem. Do that for sure.2012-05-12

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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!

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    It's ambiguous.2012-05-13
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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: $|(f(x)-K|<\frac{\epsilon}{2} \text{ and } |g(x)-L|<\frac{\epsilon}{2}$

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    @Dystopian $|f(x)+g(x)-(K+L)|$ is less than $\epsilon$ because it is less than or equal to $|(f(x)-K)|+|(g(x)-K)|$ and $|(f(x)-K)|+|(g(x)-K)|$ is less than $\epsilon$. In short, yes. (Unless I am misunderstanding you.)2012-05-13