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DDarthVader
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Homework Statement
Hello! Now I want to prove this using only δ and ε:[tex]\lim_{x\rightarrow p}f(x)+g(x)=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)[/tex]
Homework Equations
The Attempt at a Solution
These are my attempts to solve the problem.
First we start with
[tex]\lim_{x\rightarrow p}f(x)+g(x)[/tex]
if the limit do exist (and we know it does) then
[tex]p-\delta <x<p+\delta \Rightarrow f(p)-\epsilon <f(x)+g(x)<f(p)+\epsilon[/tex]
and it's limit is given by
[tex]0<x-p<\delta \Rightarrow f(x)+g(x)-L<\epsilon[/tex]
[tex]0<x-p<\delta \Rightarrow L>-\epsilon+f(x)+g(x)[/tex]
now we do the same thing for f(x)
if the limit do exit then
[tex]p-\delta <x<p+\delta \Rightarrow f(p)-\epsilon <f(x)<f(p)+\epsilon[/tex]
and it's limit is given by
[tex]0<x-p<\delta \Rightarrow f(x)-L<\epsilon[/tex]
[tex]0<x-p<\delta \Rightarrow L>-\epsilon+f(x)[/tex]
now we do it for g(x)
[tex]p-\delta <x<p+\delta \Rightarrow g(p)-\epsilon <g(x)<g(p)+\epsilon[/tex]
and it's limit is given by
[tex]0<x-p<\delta \Rightarrow g(x)-L<\epsilon[/tex]
[tex]0<x-p<\delta \Rightarrow L>-\epsilon+g(x)[/tex]
Now we see that [tex]L>-\epsilon+g(x)[/tex] and [tex]L>-\epsilon+f(x)[/tex] hence
[tex]f(x)=g(x)[/tex]
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...
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