Question

lim t->infinity of (log(t-3)-log(t^2+6t+9))

Answer 1

I think i get indeterminant form infinity-infinity, so i make it
log(t-3)/(-1/log(t^2+6t+9)??

Answer 2

no use property of logs

Answer 3

rewrite as
\[\log(\frac{t-3}{t^2+6t+9})\]

Answer 4

negative infinity.

Answer 5

Would going the other way, as I mentioned, work? Or is that headed int he wrong direction completely?

Answer 6

then
\[\lim_{t \rightarrow \infty}\frac{t-3}{t^2+6t+9}=0\] and so your limit, as sheg said, is
\[-\infty\]

Answer 7

i think you are confusing that with the method for solving
\[0\times \infty\] for example where you would rewrite the expression. but in this case you have
\[\infty - \infty\] and that is different

Answer 8

I'm quite certain infinity-infinity is an indeterminant form?

Okay so doing your method, how did you get rid of the log? and why did you set it equal to 0?

Answer 9

\[\log((t-3)/(t^2+6t+9)=\log((t-3)/(t+3)(t+3))\] But how can i simplify that?