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Prove the Sum Rule for Limits

  1. Jul 23, 2014 #1
    Prove the Sum Rule for Limits

    $$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$

    Proof

    Assume the following:
    $$\lim_{x \to a} f(x) = L, \space\lim_{x \to a} g(x) = M$$
    Then, by definition
    ##\forall \epsilon_1 > 0, \exists \delta_1 > 0## such that ##0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon_1##
    and
    ##\forall \epsilon_2 >0, \exists \delta_2 > 0## such that ##0<|x-a|<\delta_2 \implies |g(x)-M|<\epsilon_2##

    Choose ##\delta = \min(\delta_1, \delta_2)##.

    Then ##0<|x-a|<\delta \implies |f(x)-L|<\epsilon_1## and ##|g(x)-M|<\epsilon_2##.

    Notice ##|[f(x) + g(x)] - [L+M]| = |[f(x)-L] + [g(x)-M]| \leq |f(x)-L| + |g(x)-M|## by the triangle inequality. Then ##|[f(x) + g(x)] - [L+M]| \leq |f(x)-L| + |g(x)-M| < \epsilon_1 + \epsilon_2 = \epsilon##.

    Then ##|[f(x) + g(x)] - [L+M]| < \epsilon##. Therefore we conclude

    $$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$



    I was comparing my proof (the one above) to the proof offered on page 93 of James Stewart's Calculus (6e) and noticed that instead of ##\epsilon_1## and ##\epsilon_2## he uses ##\frac{\epsilon}{2}## for each of these so that their sum is ##\epsilon##.

    Is there anything incorrect with my method? I figure that my approach with two different possible values of ##\epsilon_i## for each of the limits is just a more generalized way of saying the same thing - but the other proofs I've looked at online all use ##\frac{\epsilon}{2}## - is there any reason for this?

    Thanks in advance!
     
  2. jcsd
  3. Jul 23, 2014 #2

    LCKurtz

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    If you had started with a statement of what you want to prove, you would have written:

    To prove: Given ##\epsilon > 0##, show there is ##\delta > 0## such that if ##0<|x-a|< \delta## then ##|(f(x)+g(x)) - (L+M)| < \epsilon##. And that is why you would pick the intermediate values in your argument so it all comes out less than ##\epsilon## at the end.

    Your proof should start with "Suppose ##\epsilon > 0##" Then give your argument.
     
  4. Jul 23, 2014 #3
    Oh, I see so it would be something like this:

    Proof
    Let ##\epsilon>0## be given. We must find ##\delta > 0## such that ##0<|x-a|<\delta \implies |f(x)+g(x) -(L+M)|<\epsilon##. By applying the triangle inequality we can write ##|f(x)+g(x) - (L+M)| \leq |f(x)-L| + |g(x)-M|##.

    I will pause the proof here momentarily. The proofs I have seen have continued by letting each of these terms be less than ##\frac{\epsilon}{2}## - my question is whether or not I could let one of them be less than ##\frac{\epsilon}{3}## and the other be less than ##\frac{2\epsilon}{3}## - the sum of these terms is still ##\epsilon## so this should still be valid correct? I will attempt to complete the proof using these values.

    Getting back to the main body of the proof:

    Let ##|f(x) - L|<\frac{\epsilon}{3}## and ##|g(x)-M|<\frac{2\epsilon}{3}##. Since ##\epsilon>0## it follows that both ##\frac{\epsilon}{3}>0## and ##\frac{2\epsilon}{3}>0##. Since both the limit of ##f(x)## and ##g(x)## exist as ##x## approaches ##a## there exist ##\delta_1## and ##\delta_2## satisfying the following conditions:

    $$0<|x-a|<\delta_1 \implies |f(x)-L|<\frac{\epsilon}{3}$$
    $$0<|x-a|<\delta_2 \implies |g(x)-M|<\frac{2\epsilon}{3}$$

    Choose ##\delta=\min(\delta_1, \delta_2)##. If ##0<|x-a|<\delta## then ##0<|x-a|<\delta_1## and ##0<|x-a|<\delta_2## so ##|f(x)-L|<\frac{\epsilon}{3}## and ##|g(x)-M|<\frac{2\epsilon}{3}##.

    Then ##|f(x)+g(x) - (L+M)| \leq |f(x)-L| + |g(x)-M| < \frac{\epsilon}{3} + \frac{2\epsilon}{3} = \epsilon##.

    This completes the proof.



    What was bothering me was that every proof I saw relied upon splitting the two terms equally into ##\frac{\epsilon}{2}## when I figured that it should be possible to split ##\epsilon## into two uneven terms.

    Does the above constitute a valid proof?
     
  5. Jul 24, 2014 #4

    LCKurtz

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    Yes. It doesn't matter how you break it up as long as the total parts add to less than ##\epsilon##.
     
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