Question

Limit of lnx-ln(x^2+1) as x goes to infinity

Answer 1

\[\lim_{x \rightarrow \infty}[\ln(x)-\ln(x^2+1)] \\ \lim_{x \rightarrow \infty}\ln(\frac{x}{x^2+1})\]

Answer 2

where is the inside going

Answer 3

x/(x^2+1) goes to what as x->infty?

Answer 4

oh duh

Answer 5

then just use l'hopital

Answer 6

completely forgot about that limit property

Answer 7

\[u=\frac{x}{x^2+1} \\ u \rightarrow\text{ to what ? } \text{ as } x \rightarrow \infty \]
\[\lim_{x \rightarrow \infty}\frac{x}{x^2+1}\]
you don't need l'hospital for this but you can

Answer 8

wait wouldnt the limit go to 0?

Answer 9

\[\lim_{u \rightarrow 0}\ln(u)=?\]

Answer 10

I put u->0 as x->infty
since as x->infty u=x/(x^2+1)->0

Answer 11

So then then lim as x goes to infinity IS 0

Answer 12

well that is for the inside

Answer 13

it isn't really cool to say this but what is ln(0)

Answer 14

isnt the lim of ln(x/x^2+1) equal to ln of lim of (x/x^2+1)?

Answer 15

yeah

Answer 16

sorta

Answer 17

well ln(0) goes towards negative infinity

Answer 18

right

Answer 19

I liked doing a substitution

Answer 20

ok i'll try it with a substitution but ive never done it that way before

Answer 21

it makes more sense to me
\[\text{ \let } u=\frac{x}{x^2+1} \\ \text{ since } \lim_{x \rightarrow \infty} u =\lim_{x \rightarrow \infty}\frac{x}{x^2+1}=0 \\ \text{ so we have } \lim_{u \rightarrow 0} \ln(u)=-\infty \\ \text{ so we are saying } \\ \lim_{x \rightarrow \infty}\ln(\frac{x}{x^2+1})=\lim_{u \rightarrow 0}\ln(u)=-\infty \]
But this does make sense/

Answer 22

that was suppose to end with a ?

Answer 23

-infinity