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Proof of Lemma (Limit Theorem 2)

  1. Jul 30, 2013 #1
    Hello guys, can you please help me to understand this proof of division lemma, part of the limit theorem. It is taken from Spivak's Calculus Chapter 5, page 101.

    Lemma 3
    [tex]\text{If }y_0 \neq 0\text{ and }min(\frac{|y_0|}{2}, \frac{ε|y_0|^2}{2})\ \\
    \text{ then } y \neq 0 \text{ and }|\frac{1}{y}-\frac{1}{y_0}|<ε


    We have ##|y_0|-|y| \leq |y-y_0| < \frac{|y_0|}{2}##
    (In book it is written ##\frac{y_0}{2}## I take it that is mistyped because of the subsequent line and the fact the inequality would be false.

    So ##|y| > \frac{y_0}{2}##. In particular, ##y \neq 0##, and


    I don't understand how he came to the rightmost inequality, can you guys please elaborate how he got the conclusion?

    Thank You
  2. jcsd
  3. Jul 30, 2013 #2
    Although I don't have Spivak handy, I think you probably have a typo in the first line of the Lemma. Did you mean to say |y-yo| < min(|yo|/2, ε|yo|^2/2) ?

    The rest of the proof makes sense to me. I'm not sure what you mean by 'rightmost' inequality. If you can specify which inequality you are referring to specifically, I can clarify how he arrived at it.

    Junaid Mansuri
  4. Jul 30, 2013 #3
    Yep you're right. This proof is taken from the book, it's indeed valid. Can you please elaborate the last set of inequality, after it is written thus?
  5. Jul 30, 2013 #4
    As far as the last sentence in the proof, after the Thus:

    Start by taking |1/y - 1/yo| and turn it into one fraction using the LCD, which is y·yo.
    Notice that the fraction: |yo-y|/(|y||yo|) can be rewritten as a product of three expressions as follows:
    |yo-y|/(|y||yo|) = 1/|y| · 1/|yo| · |yo-y|
    We will next replace the first and third of these three expressions by expressions that are larger than them, hence making the entire product larger:

    The first of these three is 1/|y| which will be replaced by 2/|yo| as it is larger than it.
    The second of these three is 1/|yo| which we will leave alone.
    The third of these three is |yo-y| which from the Lemma we know is less than min(|yo|/2,ε|yo|^2/2). This means that |yo-y| is less than both of them. In particular, it is less than ε|yo|^2/2. Thus, we will replace
    |yo-y| by ε|yo|^2/2 which is larger than it.

    Thus it follows that |1/y - 1/yo| < 2/|yo| · 1/|yo| · ε|yo|^2/2.
    Multiplying these three expressions and simplifiying we get:
    |1/y - 1/yo| < ε

    which was the desired end of the proof.

    Hope that makes sense.

    All the best,
    Junaid Mansuri
    Last edited: Jul 30, 2013
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