. Differential and integral calculus, an introductory course for colleges and engineering schools. + aa Oil + «2 + • • • + OLn Let the student give the proof of this. This theorem applies in the present case, and, therefore, takingthe limits in (a), we have limftslim/5i±ft+ - n = oo OtT Hence niooai+a!2+ • • • &n n = ooOCa But, by hypothesis, the first and last limits are each 1 lim-f n=oo 2/« 1, and from this it follows that lim (0i + 02 + . . + 0n) = lim (ai + a2 + . +a»). As an example of the application of this theorem, let us find thearea in polar coordinates of the sector OBC. We divide
. Differential and integral calculus, an introductory course for colleges and engineering schools. + aa Oil + «2 + • • • + OLn Let the student give the proof of this. This theorem applies in the present case, and, therefore, takingthe limits in (a), we have limftslim/5i±ft+ - n = oo OtT Hence niooai+a!2+ • • • &n n = ooOCa But, by hypothesis, the first and last limits are each 1 lim-f n=oo 2/« 1, and from this it follows that lim (0i + 02 + . . + 0n) = lim (ai + a2 + . +a»). As an example of the application of this theorem, let us find thearea in polar coordinates of the sector OBC. We divide OBC into smaller sectors, subdivide these into stillsmaller ones, and so on ad infinitum, taking care that the methodof subdividing be such that the smallsectors shall each have the limit n be the number of small sectorsafter any subdivision and OPP thetype. Since OBC is the sum of allthese n sectors, it is also the limit oftheir sum as n is indefinitely increased;that is (b) area OBC = lim % OPP. n = oo 0 In the second member of this equation we have the sum of an. 250 INTEGRAL CALCULUS §173 infinite number of positive infinitesimals, and the limit of thissum is finite. Hence, by our theorem, each of the infinitesimalsectors OPP may be replaced by another infinitesimal, providedtheir ratio has the limit 1. On each component sector OPP wenow construct, as in Art. 170, an interior circular sector shall show that OPN is the infinitesimal which may replaceOPP in (b). Construct on OPP the exterior circular sectorOPM. It is obvious that OPP OPMOPN < OPP < OPM and 1 < ^^ < —^, whence , v , < r OPP < .. OPM (c) 1 = ^opn = ^Zopn Now OPN = ip2Ad = iP2dd and OPM = \ (p + Ap) OPM (p + Ap)2OPA^ p2 and lim OPN -ft, (,+*£) -1. n=oo \ p / OPP Consequently, by reason of (c), lim npAr = 1, and, by Duhamels Theorem each OPP in (b) may be replaced by the corresponding OPAf, that is, by \ p2dd, so that we have y y area OBC = lim Y J p2 dd
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