Question

-1/5 ln(x^2+1) +2/5 ln(2x+1) as x->inf

Answer 1

I'm having some problems with the algebra here. It's supposed to be a fixed number but I can't seem to figure it out.

Answer 2

use the log properties :)

Answer 3

first factor out that 5 in the denominator

\(\Large -\ln (x^2+1) = \ln(x^2+1)^{-1}\)

Answer 4

\( 2 \ln (2x+1) = \ln [(2x+1)^2 ]\)

Answer 5

then combine those 2 'ln's

\(\ln A+\ln B = \ln AB\)

Answer 6

do this much and let me know what you get :)

Answer 7

also, this will help you in later steps :
\(\Large \lim \limits_{x\to a} \ln f(x) = \ln ( \lim \limits_{x\to a} f(x))\)

Answer 8

Ahh I see now, I never even thought of getting back that 2 as an exponent. Thank you. I got 2ln(2)/5

Answer 9

thats what i got too :)