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.(adsbygoogle = window.adsbygoogle || []).push({});

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]

QUESTION:

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?

Thanks,

Brian

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

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