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Composition goes to 0? f(x)/x -> 0 and g(x) -> 0 prove g(f(x))/x->0

  1. Nov 25, 2009 #1
    I've been looking at different math books that have analysis problems to get more perspectives on how to approach various analysis problems. I was following along in the "Derivatives" section of the book "Mathematical Thinking" by D'Angelo and West, second edition, and arrived at a lemma with corresponding proof that doesn't make sense to me.

    16.5 Lemma: Suppose e is an error function (any function such that [tex]lim_{h \rightarrow 0}\frac{e(h)}{h} = 0[/tex] ). If [tex]s(h) \rightarrow 0 [/tex] then the composition [tex]e \circ s[/tex] is an error function.

    The proof given goes like: for every [tex] \epsilon > 0 [/tex] we can choose [tex] \delta > 0 [/tex] such that [tex] |t| < \delta \Rightarrow |e(t)| \le |t| \epsilon [/tex]. Therefore [tex] |e(s(h))| \le |s(h)| \epsilon [/tex] for [tex] |s(h)| < \delta [/tex]. Since [tex] s(h) \rightarrow 0 [/tex] we can choose [tex] \delta ' [/tex] s.t. [tex] |h| < \delta ' \Rightarrow |s(h)| < \delta [/tex] and hence [tex] |h| < \delta ' \Rightarrow |e(s(h))/h| < \epsilon [/tex]. This proves that [tex] (e \circ s)(h) / h \rightarrow 0. \blacksquare [/tex]

    However, I don't follow the last step. It seems to me that we only have [tex] |h| < \delta ' \Rightarrow |e(s(h))/s(h)| \le \epsilon [/tex], how can we say this implies [tex] |e(s(h))/h| < \epsilon [/tex] ? The only way this is true would be if [tex] h > s(h) [/tex] but as s(h) is some function of h, I don't see how this could be true universally.

    I would greatly appreciate help, it's bugging me I can't figure it out.

    Maybe the proof is wrong and the error was overlooked, if so is there another way to prove it?

  2. jcsd
  3. Nov 25, 2009 #2
    |h| < \delta \Rightarrow |e(s(h))/h| < \epsilon

    Note that,

    |h| < \delta \Rightarrow |e(s(h))/h | = | e (\delta) / \delta | < \epsilon

    Which of course implies what was to be proven.
  4. Nov 25, 2009 #3
    I don't see why this is so...


    |h| < \delta \Rightarrow |e(h)/h | < \epsilon

    for that specific h, it does not imply [tex] |e(h)/d | < \epsilon [/tex] for any [tex] d < \delta [/tex], obviously this is false because we could make d arbitrarily small while fixing e(h). Similarly, [tex] |h| < \delta [/tex] and [tex] |s(h)| < \delta [/tex] in no way implies they are equal, e.g. define [tex] s(h) = h/2 \Rightarrow s(h) < \delta [/tex] if [tex] h < \delta [/tex]. In fact, even if f(x) goes to 0, this does not imply [tex] f(x) \le x [/tex] for small x, consider srqt(x) on [0,infinity). sqrt(x) goes to 0 as x goes to 0, but sqrt(x) > x for x < 1 !

    I might be misunderstanding what you are trying to say, in any case I fail to see your point.

    Note the problem I was having with understanding the proof is due to this implication:

    |h| < \delta ' \Rightarrow |e(s(h))/h| < \epsilon

    I don't see why it holds, obviously if it holds I see the proof is finished.
  5. Nov 25, 2009 #4
    An idea that may work:

    for a particular h (with abs. val. < delta and delta'), if |s(h)| <= |h| then obviously [tex] |e(s(h))/h| \le |e(s(h))/s(h)| < \epsilon [/tex]. On the other hand, if |s(h)| > |h|, then because e(x) -> 0 as x->0 if we can say e(h) > e(s(h)) then we can say [tex] |e(s(h))/h| \le |e(h)/h| < \epsilon [/tex]. However, I am not sure this holds.
  6. Nov 26, 2009 #5
    I realized my previous example (using sqrt(x)) gives a counter-example which disproves the given Lemma, i.e. the book is WRONG. I wasted all that time for a stupid error in a book... They probably meant to say s(h) is also an error function, than it would hold, just going to 0 isn't enough.

    Counter-example: take e(h) = h^2 , s(h) = sqrt(h) (or sqrt(|h|) on reals). Then e(h)/h = h ->0 as h -> 0. sqrt(h) -> 0 as h -> 0. But, e(s(h)) = h (or |h| on reals), so that e(s(h))/h -> 1 (or does not exist on reals) as h->0 => e(s(h)) is not an error function.
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