2ln(1+3n) - ln(4+n^2) = ln (3n+1)^2/(n^2+4)
now,
(3n+1)^2 / (n^2+4 ) = (9n^2+6n+1)/(n^2+4)
as n->infinity,
that -> 9n^2/n^2 = 9
so, ln(that stuff) = ln(9)
Find the limit of the sequence 2ln(1+3n) - ln(4+n^2)
What I did:
lim as n approach inf of an = 2*lim ln (1+3n)/(4+n^2)
= 2 * lim ln (1/n^2 + 3n/n^2) / (4/n^2 + n^2/n^2)
= 2 * lim ln (1/n^2 + 3/n) / (4/n^2 + 1)
= 2 * lim ln (1/inf^2 + 3/inf) / (4/inf^2 +1)
=2 * ln (0 + 0) / (0 + 1) =2* ln [ 0 / 1 ] = 2 * ln(0)
lim of ln(0) from the left does not exist
lim of ln(0) from the right is - Infinity
so lim as n approach inf of an = 2*lim ln (1+3n)/(4+n^2) = - Inf
But my teacher answer is ln(9).
I don't know whats going on...
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