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)<br />

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 &lt;x&lt;p+\delta \Rightarrow f(p)-\epsilon &lt;f(x)+g(x)&lt;f(p)+\epsilon and it's limit is given by 0&lt;x-p&lt;\delta \Rightarrow f(x)+g(x)-L&lt;\epsilon 0&lt;x-p&lt;\delta \Rightarrow L&gt;-\epsilon+f(x)+g(x)
now we do the same thing for f(x)
if the limit do exit then<br /> p-\delta &lt;x&lt;p+\delta \Rightarrow f(p)-\epsilon &lt;f(x)&lt;f(p)+\epsilonand it's limit is given by <br /> 0&lt;x-p&lt;\delta \Rightarrow f(x)-L&lt;\epsilon 0&lt;x-p&lt;\delta \Rightarrow L&gt;-\epsilon+f(x)
now we do it for g(x)<br /> p-\delta &lt;x&lt;p+\delta \Rightarrow g(p)-\epsilon &lt;g(x)&lt;g(p)+\epsilonand it's limit is given by0&lt;x-p&lt;\delta \Rightarrow g(x)-L&lt;\epsilon0&lt;x-p&lt;\delta \Rightarrow L&gt;-\epsilon+g(x)
Now we see that L&gt;-\epsilon+g(x) and L&gt;-\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 .