Question

I'm having some trouble with this limit: (3x)ln((x-1)/(x+1)) as x approaches infinity.

Answer 1

I'm trying to write to write an equation but I can't see the "equation box."

Answer 2

I just need someone to guide through a problem I stumbled upon. I'm aware that as I approach infinity I will get a indeterminate product of the type infinity*0.

Answer 3

By rewriting the limit as a quotient, I get something like lim x-> infinity of ln((x-1)/(x+1))/(1/x), in which case I easily get a 0/0 indeterminate form. Using L'hospital's rule, I differentiate the numerator and the denominator of the limit and get (2/(x^2-1))/(1/x).

Answer 4

My point is that by doing this I get a solution of 0, but according to the book, the limit should be -6.

Answer 5

rewrite it as [ ln(x-1) - ln(x+1)] / (1/(3x)), and then use L'hospital rule. You'll get -6

Answer 6

Yeah, I did that, but how can I use L'hospital's rule on an indeterminate difference of the type infinity - infinity. Isn't L'hospital's rule reserved only for 0/0?

Answer 7

Not to mention that I tried doing that but I end up with ((1/x)-(1/x))/(1/x)

Answer 8

I mean (1/(x-1)-1/(x+1))/(1/x)

Answer 9

Which still gives me 0 as the solution.

Answer 10

No, I'm getting a solution of 0, but according to my book (and wolfram), the limit is -6.

Answer 11

Is the numerator (3x)ln((x-1) and the denominator (x+1) ?

Answer 12

Oh the (x+1) is inside the ln?

Answer 13

Substitute x = 1/t. As x->inf, t->0

Answer 14

notice that ln( (x-1)/(x+1) ) approaches 0 when as x approaches 0

so you have 0/0, which now you use L'hospital rule

Answer 15

i mean x approaches infinity

Answer 16

Thank you, but I am still a little confused as to why the different results. Certainly L'hospital's should apply for something like lim x-> infinity of (2/(x^2-1))/(1/3x). Well, I'm not sure if it's that certain, but I don't see any problem besides not having the right answer, which is a huge problem.

Answer 17

When you applied L'H you did not differentiate the denominator initially. The denom should be -1/x^2.

Answer 18

d/dx 1/(3x) = -1/(3x^2),

(2/(x^2-1))/(-1/(3x^2)) will approaches -6 as x approaches infinity

Answer 19

lim x-> infinity of 3 * ln((x-1)/(x+1))/(1/x)
inf/inf. L'H can be applied:
3 * lim x_> infinity { 1/(x-1) - 1/(x+1) } / ( -1 / x^2 } =
3 * lim x-> infinity 2 / (x^2 - 1) * -x^2 =
-6 * lim x->infinity 1 / (1 - 1/x^2)
-6 * 1 = -6

Answer 20

ranga, thank you so much for pointing that out. I was repeating the same mistake every time I worked the limit out. Thanks sourwing for your help.

Answer 21

You are welcome.