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alfoguasta
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 limite esame 24 giugno 2009

qualcuno mi può risolvere questo limite??

il prof ha detto che si dovrebbe risolvere con Taylor e McLaurin..??


lim
x---->0 3x3 + x5
_____________
sin 2x - 2x cos x = -9
25-06-2009 18:19
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el-mundo
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potresti farlo anche con de l'hopital ma viene lunghissimo!!!!!!
25-06-2009 19:00
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Manuel333
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io l'ho fatto con de l'hopital risolto alla 4° derivata
26-06-2009 07:21
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alfoguasta
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se non con de hopital come si potrebbe risolvere???
26-06-2009 08:02
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alfoguasta
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qualcuno sa come risolverlo con taylor??
26-06-2009 14:18
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karplus
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Applicando Taylor viene fuori così il limite:



Facendo i conti viene proprio -9

Last edited by karplus on 29-06-2009 at 17:30
27-06-2009 18:06
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Supernick
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si, in effetti con opital ti complichi di besta :)
27-06-2009 18:10
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alfoguasta
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Originally posted by karplus
Applicando Taylor viene fuori così il limite:



Facendo i conti viene proprio -9



scusate l'ignoranza..ma come si sviluppano i calcoli per avere come risultato appunto -9??

grazie
29-06-2009 12:13
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el-mundo
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si in effetti neanche a me vengono!
29-06-2009 15:38
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el-mundo
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no no mi vengono!!!
Alfins stai attento ai segni e guarda come esce.
ciao
29-06-2009 16:33
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karplus
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Mea culpa, avevo dimenticato di inserire una potenza. Al denominatore compare (-4/3)*x^3.
Se fate i conti alla fine viene -9 -3x^2, sostituite zero ad x e ottenete -9
29-06-2009 17:32
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el-mundo
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Originally posted by karplus
Mea culpa, avevo dimenticato di inserire una potenza. Al denominatore compare (-4/3)*x^3.
Se fate i conti alla fine viene -9 -3x^2, sostituite zero ad x e ottenete -9


ma va!
veramente
acendo i conti viene sopra 3x^3
--------
2/6 x^3


ribalti la frazione sotto e ti viene 3 (moltopl.) -3 = -9
30-06-2009 08:14
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karplus
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Al posto di sin(2x) ci va 2x-(4/3)x^3, non 2x-(4/3)x come avevo scritto prima.
Il passo successivo della serie del seno é la potenza alla terza, non ancora alla prima, altrimenti viene meno il concetto di serie :D

Posto per intero lo svolgimento dell'esercizio, per essere più chiaro:



PS beh dopotutto io l'esame l'ho già passato a Gennaio, quindi mi confronto in tutta tranquillità! :D

Last edited by karplus on 30-06-2009 at 17:29
30-06-2009 12:45
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kimin@
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ma scusami karplus, il sen x non è uguale a x - x alla 3°/ 3 fattoriale??

quindi il 3! sarebbe un 6... giusto?? non riesco a capire il xkè nel secondo passaggio di quel 4/3 x alla 3° :cry::cry::cry::cry::cry:

Last edited by kimin@ on 01-07-2009 at 09:27
01-07-2009 08:19
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kimin@
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u mamma mia... ritiro la domanda idiota.... xD
01-07-2009 09:28
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