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Math Help - How to prove? (limit)

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    How to prove? (limit)

    How to prove lim(f(x)+g(x))= a+b as x->x0 and lim f(x)=a, lim g(x)=b? Could someone help me?
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    Quote Originally Posted by antero View Post
    How to prove lim(f(x)+g(x))= a+b as x->x0 and lim f(x)=a, lim g(x)=b? Could someone help me?
    proofs of various limit properties here ...

    Pauls Online Notes : Calculus I - Proof of Various Limit Properties
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    Thank you. But I didn't understand how to find out the value of delta?
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    Look at |f(x)+g(x)- (a+b)|= |(f(x)-a)+(g(x)-b)| and use the fact that |(f(x)-a)+ (g(x)-b)|\le |f(x)-a|+ |g(x)- b|. In order to make |(f(x)-a)+ (g(x)-b)|< \epsilon, we have to make sure that |f(x)-a| and |g(x)-b| add up to something less than \epsilon. One way to do that is to make sure that |f(x)-a| and |g(x)-b| each are less than \epsilon/2. Since \lim_{x\to x_0} f(x)= a, there exist some \delta_1 such that if [tex]|x- x_0|< \delta_1[\math], |f(x)- a|< \epsilon/2. Since \lim_{x\to x_0} g(x)= b, there exist some \delta_2 such that if |x- x_0|< \delta_2, |g(x)- b|< \epsilon/2.

    To be sure that both of those happen, choose \delta to be less than or equal to the smaller of \delta_1 and \delta_2. That way, if |x-x_0|< \delta, both |x-x_0|< \delta_1 and |x-x_0|< \delta_2 are true.
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