# lim (n to inf) [ln (1+3n+4n^2)-ln(5+6n+2n^2)]?

lim (n to inf) [ ln (1+3n+4n^2) - ln(5+6n+2n^2) ]

My Friend and I have trouble on this question

what I do is this:

lim (n to inf) [(1+3n+4n^2)/(5+6n+2n^2)] and then L'hopital. the answer I got is 1

My friend write is like this:

ln lim(n to inf) [(1+3n+4n^2)/(5+6n+2n^2)]. she factor n^2. and the final answer is ln(2)

We want to know what could be possibly wrong?

### 3 Answers

- 4 months ago
I'd combine them and simplify

ln((4n^2 + 3n + 1) / (2n^2 + 6n + 5)) =>

ln(n^2 * (4 + 3/n + 1/n^2) / (n^2 * (2 + 6/n + 5/n^2))) =>

ln((4 + 3/n + 1/n^2) / (2 + 6/n + 5/n^2))

n goes to infinity

ln((4 + 0 + 0) / (2 + 0 + 0)) =>

ln(4/2) =>

ln(2)

a(n) = ln(1 + 3n + 4n^2) - ln(5 + 6n + 2n^2)

n goes to infinity

a(n) = ln(inf) - ln(inf) = inf - inf

That's indeterminate, so we can use L'Hopital's rule

a(n) = ln((4n^2 + 3n + 1) / (2n^2 + 6n + 5))

e^(a(n)) = (4n^2 + 3n + 1) / (2n^2 + 6n + 5)

f(n) = 4n^2 + 3n + 1

f'(n) = 8n + 6

g(n) = 2n^2 + 6n + 5

g'(n) = 4n + 6

f'(n) / g'(n) =>

(8n + 6) / (4n + 6) =>

2 * (4n + 3) / (2 * (2n + 3)) =>

(4n + 3) / (2n + 3)

We get inf/inf, so we can use L'hopital's rule again

f(n) = 4n + 3

f'(n) = 4

g(n) = 2n + 3

g'(n) = 2

f'(n) / g'(n) =>

4/2 =>

2

e^(a(n)) = 2

a(n) = ln(2)

So what went wrong is that you did your math incorrectly. I just used L'hopital's rule and got ln(2)

- ted sLv 74 months ago
you likely made an error since L'H would yield { on the rational expression } 8 / 4 ....= 2