Notes on Homework #7Exercise 4.2.1 (c)Most of you had the right idea, writing|x3-8|=|x-2| |x2+ 2x+ 4|.Now you need to produce an upper bound for the second factor. If we commit ourselves to choosingδ≤1then|x-2|< δimplies that|x-2|<1, so that-1< x-2<11< x <37 = 12+ 2·1 + 4< x2+ 2x+ 4<32+ 2·3 + 4 = 19|x2+ 2x+ 4|<19.(It is important that we produce these boundsoutsidethe absolute value. That is becausea < bdoesnotimply|a|<|b|without a lower bound ana.) Now continuing in the usual way, you will come to the choiceofδ= min(1, /19).Exercise 4.2.8The most efficient way to do this is using Theorem 4.2.3 to relate the functional limits tosequence limits, and then appealing to the Order Limit Theorem (Theorem 2.3.4) for sequences.Some of you tried to work directly from the-δdefinition of limit. That is possible, but more difficultthan using sequences. Here is how it would go. LetLf= limx→cf(x) andLg= limx→cg(x). For any>0there existδf, δg>0 so that|f(x)-Lf|<whenever 0<|x-c|< δfandx∈A,|g(x)-Lg|<
This is the end of the preview.
access the rest of the document.