Doppellimites

Kann man Grenzwerte vertauschen, d.h. ist stets  lim xa ( lim yb f( x,y ) )= lim yb ( lim xa f( x,y ) )? MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaWGHbaabeaakmaa bmaabaWaaCbeaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadMhacqGHsg IRcaWGIbaabeaakiaadAgadaqadaqaaiaadIhacaGGSaGaamyEaaGa ayjkaiaawMcaaaGaayjkaiaawMcaaiabg2da9maaxababaGaciiBai aacMgacaGGTbaaleaacaWG5bGaeyOKH4QaamOyaaqabaGcdaqadaqa amaaxababaGaciiBaiaacMgacaGGTbaaleaacaWG4bGaeyOKH4Qaam yyaaqabaGccaWGMbWaaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIca caGLPaaaaiaawIcacaGLPaaacaqG=aaaaa@609E@

Das Beispielder Funktion f( x,y )={ x 2 y 2 x 2 + y 2  wenn x0 oder y0 0 wenn x=y=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm aabaGaamiEaiaacYcacaWG5baacaGLOaGaayzkaaGaeyypa0Zaaiqa aeaafaqaaeGabaaabaWaaSaaaeaacaWG4bWaaWbaaSqabeaaqaaaaa aaaaWdbiaaikdaaaGcpaGaeyOeI0IaamyEamaaCaaaleqabaWdbiaa ikdaaaaak8aabaGaamiEamaaCaaaleqabaWdbiaaikdaaaGcpaGaey 4kaSIaamyEamaaCaaaleqabaWdbiaaikdaaaaaaOWdaiaabccacaqG 3bGaaeyzaiaab6gacaqGUbGaaeiiaiaadIhacqGHGjsUcaaIWaGaae iiaiaab+gacaqGKbGaaeyzaiaabkhacaqGGaGaaeyEaiabgcMi5kaa icdaaeaacaaIWaGaaeiiaiaabEhacaqGLbGaaeOBaiaab6gacaqGGa GaamiEaiabg2da9iaadMhacqGH9aqpcaaIWaaaaaGaay5Eaaaaaa@62F4@
mit a=b=0 zeigt, dass das im Allgemeinen nicht geht, denn es ist lim x0 lim y0 x 2 y 2 x 2 + y 2 = lim x0 x 2 0 2 x 2 + 0 2 = lim x0 x 2 x 2 =1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIWaaabeaakmaa xababaGaciiBaiaacMgacaGGTbaaleaacaWG5bGaeyOKH4QaaGimaa qabaGcdaWcaaqaaiaadIhadaahaaWcbeqaaabaaaaaaaaapeGaaGOm aaaak8aacqGHsislcaWG5bWaaWbaaSqabeaapeGaaGOmaaaaaOWdae aacaWG4bWaaWbaaSqabeaapeGaaGOmaaaak8aacqGHRaWkcaWG5bWa aWbaaSqabeaapeGaaGOmaaaaaaGcpaGaeyypa0ZaaCbeaeaaciGGSb GaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIWaaabeaakmaalaaa baGaamiEamaaCaaaleqabaWdbiaaikdaaaGcpaGaeyOeI0IaaGimam aaCaaaleqabaWdbiaaikdaaaaak8aabaGaamiEamaaCaaaleqabaWd biaaikdaaaGcpaGaey4kaSIaaGimamaaCaaaleqabaWdbiaaikdaaa aaaOWdaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacaWG 4bGaeyOKH4QaaGimaaqabaGcdaWcaaqaaiaadIhadaahaaWcbeqaa8 qacaaIYaaaaaGcpaqaaiaadIhadaahaaWcbeqaa8qacaaIYaaaaaaa k8aacqGH9aqpcaaIXaaaaa@6C90@

aber lim y0 lim x0 x 2 y 2 x 2 + y 2 = lim y0 0 2 y 2 0 2 + y 2 = lim x0 y 2 y 2 =1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci GGSbGaaiyAaiaac2gaaSqaaiaadMhacqGHsgIRcaaIWaaabeaakmaa xababaGaciiBaiaacMgacaGGTbaaleaacaWG4bGaeyOKH4QaaGimaa qabaGcdaWcaaqaaiaadIhadaahaaWcbeqaaabaaaaaaaaapeGaaGOm aaaak8aacqGHsislcaWG5bWaaWbaaSqabeaapeGaaGOmaaaaaOWdae aacaWG4bWaaWbaaSqabeaapeGaaGOmaaaak8aacqGHRaWkcaWG5bWa aWbaaSqabeaapeGaaGOmaaaaaaGcpaGaeyypa0ZaaCbeaeaaciGGSb GaaiyAaiaac2gaaSqaaiaadMhacqGHsgIRcaaIWaaabeaakmaalaaa baGaaGimamaaCaaaleqabaWdbiaaikdaaaGcpaGaeyOeI0IaamyEam aaCaaaleqabaWdbiaaikdaaaaak8aabaGaaGimamaaCaaaleqabaWd biaaikdaaaGcpaGaey4kaSIaamyEamaaCaaaleqabaWdbiaaikdaaa aaaOWdaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacaWG 4bGaeyOKH4QaaGimaaqabaGcdaWcaaqaaiabgkHiTiaadMhadaahaa Wcbeqaa8qacaaIYaaaaaGcpaqaaiaadMhadaahaaWcbeqaa8qacaaI Yaaaaaaak8aacqGH9aqpcqGHsislcaaIXaaaaa@6E6F@ Geometrisch bedeutet dies, dass es einen erheblichen Unterschied desFunktionswerts macht, ob man sich erst zur x-Achse und dann auf dieser zum Koordinatenursprung bewegt oder erst zur y-Achse und dann auf der y-Achse zum Ursprung. Versuchen Sie sich das vorzustellen - die grafische Darstellung der Funktion hilft dabei.