# Prove the Sum Rule for Limits

1. Jul 23, 2014

### Tsunoyukami

Prove the Sum Rule for Limits

$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$

Proof

Assume the following:
$$\lim_{x \to a} f(x) = L, \space\lim_{x \to a} g(x) = M$$
Then, by definition
$\forall \epsilon_1 > 0, \exists \delta_1 > 0$ such that $0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon_1$
and
$\forall \epsilon_2 >0, \exists \delta_2 > 0$ such that $0<|x-a|<\delta_2 \implies |g(x)-M|<\epsilon_2$

Choose $\delta = \min(\delta_1, \delta_2)$.

Then $0<|x-a|<\delta \implies |f(x)-L|<\epsilon_1$ and $|g(x)-M|<\epsilon_2$.

Notice $|[f(x) + g(x)] - [L+M]| = |[f(x)-L] + [g(x)-M]| \leq |f(x)-L| + |g(x)-M|$ by the triangle inequality. Then $|[f(x) + g(x)] - [L+M]| \leq |f(x)-L| + |g(x)-M| < \epsilon_1 + \epsilon_2 = \epsilon$.

Then $|[f(x) + g(x)] - [L+M]| < \epsilon$. Therefore we conclude

$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$

I was comparing my proof (the one above) to the proof offered on page 93 of James Stewart's Calculus (6e) and noticed that instead of $\epsilon_1$ and $\epsilon_2$ he uses $\frac{\epsilon}{2}$ for each of these so that their sum is $\epsilon$.

Is there anything incorrect with my method? I figure that my approach with two different possible values of $\epsilon_i$ for each of the limits is just a more generalized way of saying the same thing - but the other proofs I've looked at online all use $\frac{\epsilon}{2}$ - is there any reason for this?

2. Jul 23, 2014

### LCKurtz

If you had started with a statement of what you want to prove, you would have written:

To prove: Given $\epsilon > 0$, show there is $\delta > 0$ such that if $0<|x-a|< \delta$ then $|(f(x)+g(x)) - (L+M)| < \epsilon$. And that is why you would pick the intermediate values in your argument so it all comes out less than $\epsilon$ at the end.

Your proof should start with "Suppose $\epsilon > 0$" Then give your argument.

3. Jul 23, 2014

### Tsunoyukami

Oh, I see so it would be something like this:

Proof
Let $\epsilon>0$ be given. We must find $\delta > 0$ such that $0<|x-a|<\delta \implies |f(x)+g(x) -(L+M)|<\epsilon$. By applying the triangle inequality we can write $|f(x)+g(x) - (L+M)| \leq |f(x)-L| + |g(x)-M|$.

I will pause the proof here momentarily. The proofs I have seen have continued by letting each of these terms be less than $\frac{\epsilon}{2}$ - my question is whether or not I could let one of them be less than $\frac{\epsilon}{3}$ and the other be less than $\frac{2\epsilon}{3}$ - the sum of these terms is still $\epsilon$ so this should still be valid correct? I will attempt to complete the proof using these values.

Getting back to the main body of the proof:

Let $|f(x) - L|<\frac{\epsilon}{3}$ and $|g(x)-M|<\frac{2\epsilon}{3}$. Since $\epsilon>0$ it follows that both $\frac{\epsilon}{3}>0$ and $\frac{2\epsilon}{3}>0$. Since both the limit of $f(x)$ and $g(x)$ exist as $x$ approaches $a$ there exist $\delta_1$ and $\delta_2$ satisfying the following conditions:

$$0<|x-a|<\delta_1 \implies |f(x)-L|<\frac{\epsilon}{3}$$
$$0<|x-a|<\delta_2 \implies |g(x)-M|<\frac{2\epsilon}{3}$$

Choose $\delta=\min(\delta_1, \delta_2)$. If $0<|x-a|<\delta$ then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$ so $|f(x)-L|<\frac{\epsilon}{3}$ and $|g(x)-M|<\frac{2\epsilon}{3}$.

Then $|f(x)+g(x) - (L+M)| \leq |f(x)-L| + |g(x)-M| < \frac{\epsilon}{3} + \frac{2\epsilon}{3} = \epsilon$.

This completes the proof.

What was bothering me was that every proof I saw relied upon splitting the two terms equally into $\frac{\epsilon}{2}$ when I figured that it should be possible to split $\epsilon$ into two uneven terms.

Does the above constitute a valid proof?

4. Jul 24, 2014

### LCKurtz

Yes. It doesn't matter how you break it up as long as the total parts add to less than $\epsilon$.