Question

limit of 3x^3+ln(x)/x+3x^2 as x approaches infinity.

Answer 1

maybe a salnt asymp?

Answer 2

?
\[\frac{3x^3+ln(x)}{x+3x^2}\]

Answer 3

slant even

Answer 4

saint asymptote, the patron saint of division

Answer 5

otherwise i think its just inf

Answer 6

\[3x^3+\ln(x)/x+2x^3\] typo'd sorry, dunno if that makes a difference.

Answer 7

y = 3x might be the slant..

Answer 8

it does lol

Answer 9

3/2

Answer 10

I'm bad at limits. Haha any help is greatly appreciated.

Answer 11

the infinity limits are gona be at the horizontal asymps if they exist

Answer 12

still infinity

Answer 13

i think the ln(x)/x^3 goes to zero...

Answer 14

oh i see it is \[\frac{3x^3+ln(x)}{x+2x^3}\] my mistake i misread it. amistre right, 3/2

Answer 15

http://www.wolframalpha.com/input/?i=lim {x+to+infinity}lnx%2Fx^3

im right lol

Answer 16

yes you are right of course. ration of leading coefficients.

Answer 17

ratio too

Answer 18

ration? now you are even typing like me ;)

Answer 19

How did you find that? How would you cancel out the natural log? I'm still kinda confused...

Answer 20

the log tends to slow down at inifinty; so the x^3 takes over and goes to 0

Answer 21

Ahh and then ratio of coefficients. Got it. Thanks!

Answer 22

a the ends; log(x) is pretty much a straight line

Answer 23

log grows slower than ANY polynomial, so you can ignore it safely

Answer 24

Right, right, I'm in compsci. We covered that. Should have remembered. Thanks so much! This was the first question I asked, how do I close this question?

Answer 25

you cant, its exists now forever in the ether of the cosmos ;)

Answer 26

That's... Unfortunate.

Answer 27

lord i hope not with all my wrong answers!

Answer 28

if you swear alot they might delete it lol