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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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What would happen if we choose $\delta_1$, where $\delta_1>\delta_2$? –  Salech Alhasov May 12 '12 at 12:12
    
@Dystopian: If you do what Salech noted, you will not go any further in the Theorem. Do that for sure. –  B. S. May 12 '12 at 12:18

2 Answers 2

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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You mean the smaller $\delta $ would also suffice for the other function? Oops, wait I haven't studied your example yet. –  Dystopian May 12 '12 at 12:25
    
Wow, that's pretty cool. I still can't think of it in general terms though, since the example was quite specific but that certainly did help. Thanks. –  Dystopian May 12 '12 at 14:41
    
Yes the smaller $\delta$ works for both. Think of it this way, if we know that $|f(x) - L| < \epsilon$ whenever $|x-a| < \delta$, then certainly $|f(x) - L| < \epsilon$ when $|x-a| < \delta/2$ or $|x-a| < \delta/10$, or indeed when $|x-a| < \gamma$ for any $\gamma < \delta$. –  Shaun Ault May 12 '12 at 15:25
    
Right. Last one though, for those who actually looked at the link, how did we get $<\frac{1}{2}\epsilon + \frac{1}{2}\epsilon $ then $=\epsilon$? –  Dystopian May 12 '12 at 15:37
    
Doesn't $\frac{1}{2} + \frac{1}{2} = 1$? –  Shaun Ault May 12 '12 at 22:01

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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Does that mean $|f(x)+g(x)-(K+L)|<|(f(x)-K)|+|g(x)-K)|$ Thus $|f(x)+g(x)-(K+L)|<\epsilon $ since $|(f(x)-K)|+|g(x)-K)|=\epsilon $ –  Dystopian May 13 '12 at 15:25
    
I mean $|(f(x)-K)|+|(g(x)-L)|<\epsilon $ Latex is making me dizzy. –  Dystopian May 13 '12 at 15:42
    
@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.) –  000 May 13 '12 at 17:37

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