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Limit Proof

  1. Jul 1, 2014 #1
    Suppose ##\lim_{x \to a} f(x) = \infty## and ##\lim_{x \to a} g(x) = c## where ##c## is a real number. Prove ##\lim_{x \to a} \big( f(x)+g(x) \big) = \infty##.

    Proof Assume ##\lim_{x \to a} f(x) = \infty## and ##\lim_{x \to a} g(x) = c##. Then, by definition:

    1) For every ##M > 0## there exists ##\delta_{1} > 0## such that ##0<|x-a|<\delta_1 \implies f(x)>M##.
    2) For every ##\epsilon >0## there exists ##\delta_{2} >0## such that ##0<|x-a|<\delta_{2} \implies |g(x)-c|<\epsilon##.

    We want to show that for every ##S>0## there exists ##\delta>0## such that ##0<|x-a|<\delta \implies f(x)+g(x)>S##.

    Choose ##\delta = \min(\delta_{1}, \delta_{2})##.
    Then ##0<|x-a|<\delta \implies f(x)>M## and Then ##0<|x-a|<\delta \implies |g(x)+c|<\epsilon \implies c -\epsilon < g(x) < c + \epsilon##.

    Then ##f(x) + g(x) > M - \epsilon + c = S## and so, with ##S = M + c - \epsilon## we conclude ##\lim_{x \to a} \big( f(x)+g(x) \big) = \infty##.


    Is this a valid proof? I'm not entirely sure if my argument is valid near the end. Any help is appreciated! Thanks.
     
  2. jcsd
  3. Jul 1, 2014 #2

    Zondrina

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    Seems legit.
     
  4. Jul 2, 2014 #3

    Fredrik

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    It's good, but there are a few things you can improve. g(x)+c appears to be a typo. Should be g(x)-c. I think it's not entirely clear what you're doing with S and M. It would add some clarity if you say "let S>0 be arbitrary" at the start of the proof, and then make it clear that you have chosen M and ε to be positive real numbers such that M-ε+c≥S (or =S). (It kind of looks like you let M and ε be arbitrary and then defined S by S=M-ε+c. That's not OK). That last line needs a "for all x such that |x-a|<δ" or a ##|x-a|<\delta~\Rightarrow~##.
     
    Last edited: Jul 2, 2014
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