Question

The Limit as x ---> infinity of ln(6x + 1) - ln(6x). Please help!!!

Answer 1

we will use l'hopitals rule here

Answer 2

but first we need to change it to a proper

Answer 3

to a proper what? Im not dividing anything here, its subtraction. I thought l'hospitals rule was for rationals.

Answer 4

are you working on l'hopitals

Answer 5

no, Im trying to find if a sequence is convergent
by taking the limit of the corresponding function.

Answer 6

But I dont know how to take this limit

Answer 7

GOT IT

Answer 8

its ZERO!

Answer 9

\[\lim_{x \rightarrow \infty}\ln(\frac{6x+1}{6x})=\lim_{x \rightarrow \infty}\ln(1+\frac{1}{6x})\]
\[=\ln(\lim_{x \rightarrow \infty}[1+\frac{1}{6x}])=\ln(1+\frac{1}{6}\cdot 0)=\ln(1)=0\]

Answer 10

lol ok

Answer 11

THANK YOU! at least I know im right.

Answer 12

it is zero from your eyeballs. there is no difference between
\[\ln(6x+1)\] and
\[\ln(6x)\], at least not for any value of x bigger than ten or so