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...
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)
To find the correct limit of the sequence 2ln(1+3n) - ln(4+n^2), let's go through the steps again using a different approach.
First, rewrite the expression using logarithmic properties:
2ln(1+3n) - ln(4+n^2)
= ln((1+3n)^2) - ln(4+n^2)
= ln((1+3n)^2 / (4+n^2))
Now, let's evaluate the limit as n approaches infinity. We can do this by considering the largest power of n in the numerator and denominator.
Looking at the expression (1+3n)^2 / (4+n^2), as n approaches infinity, the dominant term in the numerator is (3n)^2 = 9n^2, and the dominant term in the denominator is n^2.
So, we can rewrite the expression as:
ln((1+3n)^2 / (4+n^2))
= ln((9n^2 / n^2))
= ln(9)
Therefore, the correct limit of the sequence is ln(9), which matches the answer provided by your teacher. It seems that there was a mistake in your previous approach.