Question

lim x-> 1- ln(2x) / ln x

Answer 1

i did (1/lnx)/ (1/ln2x) to get 0/0.. dont know if that's how u should start off with

Answer 2

directly put x=1
because u can't use L'Hopitals here.

Answer 3

but can't I get 0/0 when i rearrange it the way i did?

Answer 4

ln(2x) ---> ln(2) is not 0

Answer 5

or is it that base of log is 2 ? O.o

Answer 6

isn't L'hopital is first trying to rearrange the original function to get indeterminate form

Answer 7

no it is ln(2x)

Answer 8

then 1/ln x over 1/ ln 2x will be 0/0 isnt it

Answer 9

yes, but u don't get the indeterminate form 0/0 here,

Answer 10

no, how is it 0/0 ?

Answer 11

oh..is it not.. oh yea..cuz x is not going to infinity

Answer 12

so if u cannot get indeterminate form ,u just directly sub the x in?

Answer 13

yes!

Answer 14

but the answer suppose to be -infinity

Answer 15

if i just sub it in, then ln x goes to 0

Answer 16

and it is, did u put x=1 ?

Answer 17

so denominator =0 , right ?

Answer 18

yea

Answer 19

actually, the reasoning goes like this...
since x-> 1- , x is very very near to 1, but less than 1.
so ln (1-) is very very near to 0, but less than 0.(negative)
so u get -inf, instead of + inf

Answer 20

then how about ln(2x), is it -ve infinity as well?

Answer 21

when u put x=1, its just ln 2 = 0.3

Answer 22

so ln2 / -ve infinity is still -ve infinity?

Answer 23

0.3/(-very small number) = -very bid number

Answer 24

*big

Answer 25

ln(1-) is not -inf, u can say -0

Answer 26

\(0^-\)

Answer 27

oh ok, all right. thanks for ur help again! i've done enuff math problems today :D hope u see u again later haha :)

Answer 28

sure, welcome ^_^