Calcolo del limite
1A) limn→+∞ n•ln(n+1⁄n+2) = forma ind. 0•∞
limn→+∞ ln(n+1⁄n+2) ⁄ 1⁄n = forma ind [0⁄0]
N. = limn→+∞ n+2⁄n+1 - [(n+2) -(n+1)⁄(n+2)2] == -1⁄n2
= limn→+∞ n+2 [1⁄(n+2) - (n+1)⁄(n+2)2] ⁄ n+1 == -1⁄n2
= limn→+∞ 1 - n+1⁄n+2⁄n+1 • -n2 == limn→+∞ n+2 - (n+1)⁄n+2 • -n2
= limn→+∞ 1⁄n+2 ⁄ 1⁄n+1 • -n2
= limn→+∞ 1⁄(n2 • n+2) == limn→+∞ 1⁄(n2 + 2n + n + 2)
= limn→+∞ n2⁄n2 + 3n + 2 == n2 • (-1)⁄n2 (1 + 3n⁄n2 + 2⁄n2) = -1
11) limn→+∞ n ln ( n+1⁄n+2 ) = forma ind. 0 · ∞
limn→+∞ ln ( n+1⁄n+2 ) ⁄ 1/n = forma ind [∞⁄∞]
limn→+∞ n+2⁄n+1 ( (n+2)-(n+1)⁄(n+2)2 ) == -1⁄n2
= limn→+∞ n+2⁄n+1 ( (n+2)-(n+1)⁄(n+2)2 ) ⁄ 1/n+1
== limn→+∞ 1 - n+1⁄n+2 . n2 == limn→+∞ n+2-(n+1)⁄n+2 . n2
== limn→+∞ 1⁄n+2 . n2 == limn→+∞ 1⁄(n2 + n + 2)
== limn→+∞ n2⁄(n2+3n+2) == n2-1⁄n2(1+3⁄n + 2⁄n2) =-1
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Geotecnica - Esercizi
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Esercizi Testi limiti, successioni
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