Question

How do I change the form of this case to apply L'hosptial's rule?
lim (ln|x-1| - ln|x+1|)
x->+inf
It seem to approach inf - inf, which is indefinite. In the cases I've seen this, they're fractions, and combining them does the job, but what about a case like this? Oh, and it apparently approaches 0.

Answer 1

Ohhh! I can actually force them into fractions using this;
1/a^-1 + 1/b^-1 = (a^-1 + b^-1)/a^-1 * b^-1

All of this miss to put it in factor form, omg. Is there no faster way?

Answer 2

your concerned limit is :
\[\lim_{x\to \infty } \mid { \frac{x-1}{x+1} } \mid \]

Answer 3

forgot ln of | x-1/x+1|

Answer 4

using log property you can make it condensed
then try to think how you would apply L'hopital

Answer 5

.............
Wow. That's simply Magnificent. ln|x| +- ln|y| = ln|x*/y|.
I thought of everything but that!

Answer 6

but it won't be that simple
to use l'hopital f and g must be differentiable
and abs value is not differentiable
so think of a way to approach this issue

Answer 7

isn't it possible to do this;
e^y = lim of d-1/d+1
x-> +inf
that way i finally get inf/inf

Answer 8

well you had a good idea, however, before that you need to get rid of abs values
they are causing trouble

Answer 9

let me remind you that f(x)=|x|
f' does not exist

Answer 10

now you surely you are facing the same issue here |f(x)/g(x)|

Answer 11

ln is not a problem differentiable everywhere for x>0

Answer 12

Cant' I break it into two separate cases for
x>=0 and x<0 ?

Answer 13

hmm but think again if x-1<0 you can't allow negative in ln
x+1<0 same thing

Answer 14

what I'm saying is that abs values are just tricking you
because of x+1 to be in log it has to be >0
so just get rid of abs value and you have a nice limit

Answer 15

Oh I actually meant break the absolute value function itself into a piecewise function:
x {x>=0}
-x {x<0}

Answer 16

i just wanted to caution you about this abs values

Answer 17

x here is approaching positive values so x+1>0

Answer 18

same goes for x-1

Answer 19

yes but how can I actually get rid of the absolute value,
I know for y = ln|x|
e^y = x but only as long as x is positive.

Answer 20

just remove no harm hhh

Answer 21

because your quotient is positive! it must be since negative values can't be in log

Answer 22

well i guess im confusing you hhh
the fact that x>0 (x goes to +infinity)
then x-1 must be positive
x+1 also

Answer 23

x-1 / x+1
I see. so for every negative value I in fact come out with a positive value!

Answer 24

as long as that number is not in the [-1,1] range, though.

Answer 25

What you're saying makes sense, x is approaching infinity, but then, as it "approaches" did it not at one time cross the forbidden zone {-1,1}, wouldn't that make the limit break since the whole thing is not continuous? I guess it doesn't matter since limits are about values very close to the limit point, so it's safely ignored?

Answer 26

Thank you very much, limit guy!