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.dsy:it. : Powered by vBulletin version 2.3.1 .dsy:it. > Didattica > Corsi G - M > Matematica del continuo > [com dig] Limiti: esercizi ! ! !
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Alis
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Anche a me viene 1/7!!!!
02-05-2003 19:50
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Lunik
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cmq è brutto leggere sul foglio delle soluzioni che il risultato di questo limite sia 4/11.... così uno che non si confronta con altri che fa? Impazzisce per ore e ore...mentre la soluzione giusta è un'altra????? bah!!

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02-05-2003 19:52
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valeria
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la formula per risolvere le eq. di 2' grado la ricordo... ma perchè usarla se ho radici frazionarie
quando posso scomporre il denominatore così?

(2x^2 -x -6)=(2x^2 +2x -3x -6)=[2x(x+1) -3(x+1)]=(2x-3)(x+1)
02-05-2003 20:16
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Lunik
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xè la formula è quella che si usa spesso...e al liceo è l'unica cosa che si usa... almeno così era per me...,lasciamo il topic per gli esercizi e soluzioni...grazie

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02-05-2003 20:22
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valeria
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es. lim x--> + infinito [(x+1)^(1/2) - x^(1/2)]x^(1/2)

le sol :1/2 ...a me viene 1

ok mi è riuscito..grazie

Last edited by valeria on 03-05-2003 at 14:27
03-05-2003 12:16
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Barone
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Qualcuno potrebbe darmi una mano su seguente limite:

Lim x->1+ modulo di(x-1)
-----------------
log x
Secondo me si parte dal concetto che il limite si può eliminare in quanto l'1+ è presuppone che sia sempre positivo l'argomento del modulo stesso.

Hell..p
03-05-2003 12:57
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Lunik
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 risultato lim di Barone

dà per caso 1?? sai la soluzione?

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03-05-2003 12:59
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Barone
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si
03-05-2003 13:12
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Lunik
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 risoluzione del lim di Barone

Allora io ho fatto:
modulo x - 1 è cmq sempre positivo, avendo la x che tende a +1, il modulo di (x-1) sarà un numero poco più grande di 1...
Log x = metti l'1 al posto dell'x...nel grafico... la funzione tende all'1...
Hai alla fine 1/1... giusto? O canno?

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03-05-2003 13:15
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Barone
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canni..
log di 1 è zero...non uno!!!credevo fosse infinito la sol invece da uno...boh..

vabbè vado allo stadio... a stesera.
03-05-2003 13:18
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Lunik
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si si... ho sbagliato a ragionare sul grafico... mea culpa... :(

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03-05-2003 13:21
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valeria
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qualcuno può dirmi come risolvere questo limite grazie?
lim (x--> 0+) di log [ ( x^4 + 3x^2 ) / (x -x^7 ) ]
..io non riesco ad eliminare la forma indeterminata 0/0
03-05-2003 14:27
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GinoPilotino
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Originally posted by valeria
qualcuno può dirmi come risolvere questo limite grazie?
lim (x--> 0+) di log [ ( x^4 + 3x^2 ) / (x -x^7 ) ]
..io non riesco ad eliminare la forma indeterminata 0/0

per caso viene +infinito?
03-05-2003 15:05
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valeria
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lim (x--> 0+) di log [ ( x^4 + 3x^2 ) / (x -x^7 ) ]=
sol: - infinito
come hai fatto? grazie
03-05-2003 15:26
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GinoPilotino
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allora, al numeratore domina x^4 mentre al denominatore domina -x^7. Di conseguenza il denominatore è di ordine superiore rispetto a numeratore quindi la soluzione viene infinito. Il segno poi lo ricavi dal segno della x dominante quindi - infinito.

Possibile che mi vengono solo gli esercizi degli altri?? :? :D
03-05-2003 16:40
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