- #1
ciubba
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Homework Statement
Prove that [tex]\lim_{x->a}[f(x)+g(x)]=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)][/tex]
Homework Equations
Epsilon/delta definition
The Attempt at a Solution
The book says:
Let the limit of f(x)=L and the limit of g(x)=M. Then,
[tex]\mid f(x)-L \mid<\frac{\epsilon}{2}[/tex] whenever [tex]0<\mid x-a \mid<\delta_{1}[/tex]
and
[tex]\mid g(x)-M \mid<\frac{\epsilon}{2}[/tex] whenever [tex]0<\mid x-a \mid<\delta_{2}[/tex]
Let [tex] \delta=min \big\{ \delta_{1} ,\delta_{2} \big\}[/tex] and suppose [tex]0<\mid x-a \mid<\delta[/tex]. Because [tex]\delta\leq\delta_{1}[/tex], it follows that [tex]0<|x-a|\delta_{1}[/tex] and [tex]|f(x)-L<\frac{\epsilon}{2}[/tex].
Similarly, [tex]\delta\leq\delta_{2}[/tex], it follows that [tex]0<|x-a|\delta_{2}[/tex] and [tex]|g(x)-L<\frac{\epsilon}{2}[/tex]. Therefore,
[tex]|[f(x)+g(x)]-(L+M)|<\epsilon[/tex], which implies that [tex]]\lim_{x->a}[f(x)+g(x)]=L+M=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)][/tex]
My biggest question is from where they obtained a value of epsilon/2 from. I feel that I cannot even begin to understand the proof until I understand that point.
Edit: The latex didn't come out as clearly as I'd hoped, so here is a picture: http://s9.postimg.org/7gfz3owul/Proof.gif
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