# Proving Limit Laws

1. Jan 14, 2015

### ciubba

1. The problem statement, all variables and given/known data
Prove that $$\lim_{x->a}[f(x)+g(x)]=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)]$$

2. Relevant equations
Epsilon/delta definition

3. The attempt at a solution

The book says:

Let the limit of f(x)=L and the limit of g(x)=M. Then,

$$\mid f(x)-L \mid<\frac{\epsilon}{2}$$ whenever $$0<\mid x-a \mid<\delta_{1}$$
and
$$\mid g(x)-M \mid<\frac{\epsilon}{2}$$ whenever $$0<\mid x-a \mid<\delta_{2}$$

Let $$\delta=min \big\{ \delta_{1} ,\delta_{2} \big\}$$ and suppose $$0<\mid x-a \mid<\delta$$. Because $$\delta\leq\delta_{1}$$, it follows that $$0<|x-a|\delta_{1}$$ and $$|f(x)-L<\frac{\epsilon}{2}$$.

Similarly, $$\delta\leq\delta_{2}$$, it follows that $$0<|x-a|\delta_{2}$$ and $$|g(x)-L<\frac{\epsilon}{2}$$. Therefore,

$$|[f(x)+g(x)]-(L+M)|<\epsilon$$, which implies that $$]\lim_{x->a}[f(x)+g(x)]=L+M=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)]$$

My biggest question is from where they obtained a value of epsilon/2 from. I feel that I cannot even begin to understand the proof until I understand that point.

Edit: The latex didn't come out as clearly as I'd hoped, so here is a picture: http://s9.postimg.org/7gfz3owul/Proof.gif

Last edited: Jan 14, 2015
2. Jan 14, 2015

### ehild

Read the first sentence after SOLUTION in the picture: "Assume that ε > 0 is given " You can require that both functions deviate from their limit less than half of the given ε.

3. Jan 14, 2015

### ciubba

Yes, I saw that, but I also could have defined it as < ε/3 or 2piε/4. What is the significance of making it ε/2? I looked at a proof for the product limit laws and saw a similar technique employed.

4. Jan 14, 2015

### ehild

You have to find δ so as the deviation of the combined function f(x)+g(x) from its limit is less than ε. The simplest is to choose equal ε1 and ε2 for the functions f and g, so as ε1+ε2 = ε. But you have the freedom to choose ε/3 for one function and 2/3 ε for the other one, or anything else if they are positive and their sum is ε.

5. Jan 14, 2015

### Ray Vickson

It really does not matter. For any (small) $\epsilon' > 0$ we can find $\delta > 0$ such that $|f(x)-L| < 3 \epsilon'$ and $|g(x) -M |< 2 \pi \epsilon'$ for $|x - a| < \delta$. Then we have
$$|f(x) + g(x) - (L+M)| < (3 + 2 \pi) \epsilon'$$
for $|x-a| < \delta$. Now just put $\epsilon = (3 + 2 \pi) \epsilon'$. Given any (small) $\epsilon > 0$ we can thus find $\delta > 0$ such that $|f(x) + g(x) - (L+M)| < \epsilon$ if $|x-a| < \delta$. We do that by putting $\epsilon' = \epsilon/(3 + 2 \pi)$ and using the $\delta$ that goes along with $\epsilon'$. It is just a lot neater to put $\epsilon' = \epsilon/2$.