Proving Limits with Δ & ε | Homework Statement

[DDarthVader]

Homework Statement

Hello! Now I want to prove this using only δ and ε:$$\lim_{x\rightarrow p}f(x)+g(x)=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)$$

Homework Equations

The Attempt at a Solution

These are my attempts to solve the problem.
First we start with
$$\lim_{x\rightarrow p}f(x)+g(x)$$
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...

 

[SteveL27]

Homework Statement

Hello! Now I want to prove this using only δ and ε:$$\lim_{x\rightarrow p}f(x)+g(x)=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)$$

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.

 

[DDarthVader]

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

 

[SammyS]

Homework Statement

Hello! Now I want to prove this using only δ and ε:$$\lim_{x\rightarrow p}f(x)+g(x)=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)$$

Homework Equations

The Attempt at a Solution

These are my attempts to solve the problem.
First we start with $$\lim_{x\rightarrow p}f(x)+g(x)$$ 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...

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 ε_f = ε/2 and ε_g = ε/2 .

You then know that there is some δ_f and δ_g .

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

 

[DDarthVader]

Why εf = ε/2 and εg = ε/2?

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

My δ would be δ=min{δf,δg}, correct?

 

[SteveL27]

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

Yes. I already took that class!

 

[SammyS]

Why εf = ε/2 and εg = ε/2?


My δ would be δ=min{δf,δg}, correct?

That's the correct δ.

Put it all together to see why ε_f = ε/2 and ε_g = ε/2 .