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By M. H. Protter C. B. Morrey Jr.

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X) = Lj' i = I, 2, ... ,n. ;(x). 5. 6 (Limit of a product). Suppose that lim fl (x) = LI x~a Define g(x) and lim f2 (x) = L 2. x~a = fl(x) ·fix). 2: Theorems on limits Suppose that f such that PROOF. > 0 is given. Ig(x) - LIL21 0 whenever 0 < Ix - al <~. 2) to get I g(x) - LIL21 < IL21·lfl (x) - LII + IfI (x)I'lf2 (x) < IL21'fl + M'f2' L21 Select fl = f/2L2 and f2 = f/2M. The quantities ~I and ~2 are those which correspond to the values of fl and f2' respectively.

If k > 3 and k! >(k+ 1)·2k >4·2k =2·2k +1 >2k +1 so that (k + 1) E S. Thus (k + 1) is in S whenever k is. Therefore S is inductive and so consists of all natural numbers. If n is any natural number > 3, it follows that n! > 2n. Note that in this example the inequality n! > 2n is false for n = 1, 2, 3. This made necessary the strange definition of our set S. Remark. 22. A set S of real numbers has the modified inductive property if (1) S has a smallest number and (2) (x + 1) E S whenever xES. 22 becomes: if S is a set of natural numbers with the modified inductive property, the S contains all the natural numbers greater than the smallest natural number in S.

Now, sincef(x)~oo as x~a, it follows that for any number A > I + ILl there is a 8 1 such thatf(x) is defined for 0 < Ix - al < 8 1 and If(x)1 >A for 0 < Ix - al < 81• Thus for any value x( =1= a) closer to a than both 8 and 8 1 we have the impossible situation: I + ILl < A < If(x)1 < I + ILl· 0 Remark. 16 with + 00 or - 00 instead of 00 and with one-sided limits instead of the two-sided limit require only obvious modifications. 15 on the limit of a composite function takes a somewhat different form for infinite limits.

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