LanHikari22
  • LanHikari22
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.
Mathematics
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chestercat
  • chestercat
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LanHikari22
  • LanHikari22
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?
xapproachesinfinity
  • xapproachesinfinity
your concerned limit is : \[\lim_{x\to \infty } \mid { \frac{x-1}{x+1} } \mid \]
xapproachesinfinity
  • xapproachesinfinity
forgot ln of | x-1/x+1|

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xapproachesinfinity
  • xapproachesinfinity
using log property you can make it condensed then try to think how you would apply L'hopital
LanHikari22
  • LanHikari22
............. Wow. That's simply Magnificent. ln|x| +- ln|y| = ln|x*/y|. I thought of everything but that!
xapproachesinfinity
  • xapproachesinfinity
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
LanHikari22
  • LanHikari22
isn't it possible to do this; e^y = lim of d-1/d+1 x-> +inf that way i finally get inf/inf
xapproachesinfinity
  • xapproachesinfinity
well you had a good idea, however, before that you need to get rid of abs values they are causing trouble
xapproachesinfinity
  • xapproachesinfinity
let me remind you that f(x)=|x| f' does not exist
xapproachesinfinity
  • xapproachesinfinity
now you surely you are facing the same issue here |f(x)/g(x)|
xapproachesinfinity
  • xapproachesinfinity
ln is not a problem differentiable everywhere for x>0
LanHikari22
  • LanHikari22
Cant' I break it into two separate cases for x>=0 and x<0 ?
xapproachesinfinity
  • xapproachesinfinity
hmm but think again if x-1<0 you can't allow negative in ln x+1<0 same thing
xapproachesinfinity
  • xapproachesinfinity
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
LanHikari22
  • LanHikari22
Oh I actually meant break the absolute value function itself into a piecewise function: x {x>=0} -x {x<0}
xapproachesinfinity
  • xapproachesinfinity
i just wanted to caution you about this abs values
xapproachesinfinity
  • xapproachesinfinity
x here is approaching positive values so x+1>0
xapproachesinfinity
  • xapproachesinfinity
same goes for x-1
LanHikari22
  • LanHikari22
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.
xapproachesinfinity
  • xapproachesinfinity
just remove no harm hhh
xapproachesinfinity
  • xapproachesinfinity
because your quotient is positive! it must be since negative values can't be in log
xapproachesinfinity
  • xapproachesinfinity
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
LanHikari22
  • LanHikari22
x-1 / x+1 I see. so for every negative value I in fact come out with a positive value!
LanHikari22
  • LanHikari22
as long as that number is not in the [-1,1] range, though.
LanHikari22
  • LanHikari22
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?
LanHikari22
  • LanHikari22
Thank you very much, limit guy!

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