Question

ln(x)*ln(x-1)

Suppose you wanted to prep this fcn for L'Hopital's Rule. The expression is the limit as x approaches 1 from the right side of the above fcn.

How do we do so?
ln(x)*ln(x-1)

Suppose you wanted to prep this fcn for L'Hopital's Rule. The expression is the limit as x approaches 1 from the right side of the above fcn.

How do we do so?
@Mathematics

Answer 1

\[\lim_{x \rightarrow 0+}\ln(x)\ln(x-1)\]

Answer 2

forget it

Answer 3

you cannot take the limit as x approaches 0 of ln(x-1)

Answer 4

domain
\[\ln(x-1)\] is
\[x>1\]

Answer 5

and if it is
\[\lim_{x\rightarrow 1^+}\ln(x)\ln(x-1)\] l\hopital would not apply because this is of the form
\[1\times -\infty\]

Answer 6

I think he made a typo. He wants the expression is the limit as x approaches 1 from the right side of the above fcn.

Answer 7

ok then it is not "undetermined" or whatever they say

Answer 8

limx->0 ln(x).ln(x-1) = limx->0 ln(x)./1/ln(x-1) <- use L'Hopital's rule now
limx->0 1/x/1/1/(x-1) = limx-> (x-1)/x <-use L'Hopital's rule again
limx->0 (1)/1) = 1

Answer 9

How do we make it an indeterminate form?

Answer 10

you would make a mistake to write it as a fraction say, and use l'hoptital because that only works for undetermined limits. like for example
\[\frac{0}{0}\] or
\[\frac{\infty}{\infty}\] or
\[1^{\infty}\] etc

Answer 11

ln(x) / 1/ln(x-1) yields 0/infinity, and this isn't in L'H form...

Answer 12

hold the phone
you do not " make it an indeterminate form"

Answer 13

it either is in one or it is not to begin with

Answer 14

So what the heckk do I do for this case?

Answer 15

lets go slow.

Answer 16

what is the actual problem?

Answer 17

QRAwarrior, check my answer again :)

Answer 18

if i want the limit as x goes to 0 of 1/x, i do not "change the form" that limit does not exist, so i cannot make it exist by changing the question

Answer 19

The problem is this:
Limit as x approaches 0 from the right side of (lnx)(ln(x-1))

Answer 20

are you sure

Answer 21

is this a course in complex analysis?

Answer 22

send me a medal later :) brb

Answer 23

what you are thinking of is how to change something like
\[x\ln(x)\] into
\[\frac{\ln(x)}{\frac{1}{x}}\]so that i can turn an indermined limit like
\[\lim_{x\rightarrow 0}x\ln(x)\] which is of the form
\[0\times -\infty\] into
\[\frac{\ln(x)}{\frac{1}{x}}\] which is of the form
\[\lim_{x\rightarrow 0^+}\frac{-\infty}{\infty}\] that is just algebra

Answer 24

but if your limit is not undetermined to begin with, you cannot change it into something that is

Answer 25

maybe i am not sure of the actual question, so let me be quiet

Answer 26

@Q, you keep saying x->0 but the original post says as x->1 (which makes much more sense)

for lim x->1 we know ln(x) approaches 0, so we might re-write the expression as
ln(x-1)/ (1/ln(x))

Answer 27

ooooooooooh ok that you can do. same gimmick as before as phi wrote