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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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Barone
non cambio mai avatar

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Originally posted by Lunik
già solo che NON bisogna dimenticarsi del 2 del 2x^2!!!!!



Quindi qnd voglio scomporre un polinomio che ha un numero davanti alla x^2 devo metterlo davanti alla scomposizione in radici?Questa mi è nuova...io ho fatto il solito metodo senza togliere il 2 e il limite mi risulta uguale a 2/7...lunik(o altri) illuminami...
04-05-2003 11:19
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Lunik
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trova le soluzioni...scomponi l'equazione...e senza il due fai la prova..ti viene uguale? Se si, allora sbaglio io a mettere il 2.... ma credo che non ti verrà uguale...proprio xè manca il 2! ;)

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04-05-2003 11:21
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Barone
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ok..grazie...perchè questa è una novità per me....lunik non è che hai gli appunti dell'ultima lezione da mandarmi?
04-05-2003 11:31
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Lunik
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erm no xè nn seguo...e neanche devo fare l'esame :)
purtroppo nn so chi potrebbe darteli..... :pensa:

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04-05-2003 11:35
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Elex
la zia perfetta

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barone, di che lezione ti servono? dimmi la data, sei del primo turno giusto?
04-05-2003 11:37
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Barone
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si dell'ultima lezione e quella della parità di funzione..turno uno..
aleferra@caltanet.it
04-05-2003 11:39
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Lunik
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Barone... hai visto la mia spiegazione sulla parità? NOn ti è ancora chiara??? :(

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04-05-2003 11:39
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Elex
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secondo i miei appunti l'ultima lezione è sulle funzioni continue...se ti serve cmq fammelo sapere
04-05-2003 11:43
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Barone
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chiara ma se mi mandate anche gli appunti è meglio...Grazie lunik sei preziosissima..
04-05-2003 11:43
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Barone
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I miei appunti arrivano fino agli asintoti..se hai qlc dopo mandali per favore..
grazie
04-05-2003 11:45
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Elex
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continuiamo via pm
04-05-2003 11:47
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Joda
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Originally posted by dilix
Il cosx-1/x = 0 è un limite notevole?


Limite notevole

1-cos(x)
---------- = 1/2
x^2

Cos(x) -1
----------- = - 1/2
x^2


Vai qui

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04-05-2003 11:48
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Joda
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Originally posted by valeria
Lim x->1+ |x-1|/log x

non sono sicura di quanto scrivo..ma..farei così...
..adesso è corretto?

lim (t->0+) (t)/log (t+1)=1
lim notevole lg(x+1)/x=1 per x->0
1


Si Valeria e' giusto, pero' attenta... e' t->0 non x

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04-05-2003 11:51
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valeria
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grazie joda
04-05-2003 15:20
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valeria
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es lim (x->0) [lg(1+x) + e^x -cosx] / tgx
io non riesco ad uscire dalla forma indeterminata 0/0? se riuscite a girarlo in qualche modo ?....GRAZIE
04-05-2003 15:25
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