oblicz ranice ciągu:
lim
( pierw. z n^2+ 9n kon.pierw. -n)
n do niesk.
podrzebuje na juz pomocy!!!
Stosujemy wzór:
a - b = ( a^2 - b^2) / ( a + b)
Mamy
an = [p( n^2 + 9n) - n ] = ( n^2 + 9n - n^2)/ [ p( n^2 + 9n) + n ] =
= 9n / [ p( n^2 + 9n ) + n ] = 9 / [ p( 1 + 9/n) + 1 ]
lim an = 9/2 = 4,5 , bo 9/n --> 0, gdy n --> oo
n --> oo
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Stosujemy wzór:
a - b = ( a^2 - b^2) / ( a + b)
Mamy
an = [p( n^2 + 9n) - n ] = ( n^2 + 9n - n^2)/ [ p( n^2 + 9n) + n ] =
= 9n / [ p( n^2 + 9n ) + n ] = 9 / [ p( 1 + 9/n) + 1 ]
lim an = 9/2 = 4,5 , bo 9/n --> 0, gdy n --> oo
n --> oo