# How to prove this?

1. Apr 23, 2012

1. The problem statement, all variables and given/known data
Hello! Now I want to prove this using only δ and ε:$$\lim_{x\rightarrow p}f(x)+g(x)=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)$$

2. Relevant equations

3. The attempt at a solution
These are my attempts to solve the problem.
$$\lim_{x\rightarrow p}f(x)+g(x)$$
if the limit do exist (and we know it does) then
$$p-\delta <x<p+\delta \Rightarrow f(p)-\epsilon <f(x)+g(x)<f(p)+\epsilon$$
and it's limit is given by
$$0<x-p<\delta \Rightarrow f(x)+g(x)-L<\epsilon$$
$$0<x-p<\delta \Rightarrow L>-\epsilon+f(x)+g(x)$$

now we do the same thing for f(x)
if the limit do exit then
$$p-\delta <x<p+\delta \Rightarrow f(p)-\epsilon <f(x)<f(p)+\epsilon$$
and it's limit is given by
$$0<x-p<\delta \Rightarrow f(x)-L<\epsilon$$
$$0<x-p<\delta \Rightarrow L>-\epsilon+f(x)$$

now we do it for g(x)
$$p-\delta <x<p+\delta \Rightarrow g(p)-\epsilon <g(x)<g(p)+\epsilon$$
and it's limit is given by
$$0<x-p<\delta \Rightarrow g(x)-L<\epsilon$$
$$0<x-p<\delta \Rightarrow L>-\epsilon+g(x)$$

Now we see that $$L>-\epsilon+g(x)$$ and $$L>-\epsilon+f(x)$$ hence
$$f(x)=g(x)$$
Now I'm stuck! I mean, I don't know if what I did is correct so I don't think I should go on...

Last edited: Apr 23, 2012
2. Apr 23, 2012

### SteveL27

I think they want you to prove that

$$\lim_{x\rightarrow p}(f(x)+g(x))=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)$$

Big difference.

3. Apr 23, 2012

Yes! Sorry my fault.
Anyway, do you know how to prove it?

4. Apr 23, 2012

### SammyS

Staff Emeritus
I assume that what need to do is the following.
If $\displaystyle \lim_{x\to \ p}f(x)$ and $\displaystyle \lim_{x\to\ p}g(x)$ exist, then prove that $\displaystyle \lim_{x\to\ p}(f(x)+g(x))=\lim_{x\to\ p}f(x)+\lim_{x\to\ p}g(x)\ .$​
Given ε > 0, use εf = ε/2 and εg = ε/2 .

You then know that there is some δf and δg .

What should your δ be for the limit of the sum ?

5. Apr 23, 2012

Why εf = ε/2 and εg = ε/2?

My δ would be δ=min{δf,δg}, correct?

6. Apr 23, 2012

### SteveL27

Yes. I already took that class!

7. Apr 23, 2012

### SammyS

Staff Emeritus
That's the correct δ.

Put it all together to see why εf = ε/2 and εg = ε/2 .