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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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