How to Solve a Limit Question with an Infinite - Infinite Equation

[mad]

EDITED THE EQUATION


Hello all,

I have this problem I can't solve.. it is a infinite - infinite. I tried it around 5 times and can't find the correct answer (infinite). I'm pretty sure I have to put in evidence x^2 and use a limit law but I can't find the answer.. can someone help me for this problem:

lim x -> +inf. $$x - \ln(x^2-1)$$

Thanks very much in advance.

 

[Justin Lazear]

The limit evaluates to -0.5? How do you know this?

--J

 

[mad]

The limit evaluates to -0.5? How do you know this?

--J

That's the answer in the book =)

Sorry, it was x - (lnx^2-1)

 

[Justin Lazear]

x^2 increases much faster than ln(x^2 -1). The limit should be unbounded.

The technique I'd use to evaluate it would be to write x^2 as [itex]\ln{\left(e^{x^2}\right)}[/tex] and then combine the logs. This approach gives a result of infinity, as well.

--J

 

[mad]

x^2 increases much faster than ln(x^2 -1). The limit should be unbounded.

The technique I'd use to evaluate it would be to write x^2 as [itex]\ln{\left(e^{x^2}\right)}[/tex] and then combine the logs. This approach gives a result of infinity, as well.

--J

I edited the equation. It was a typo.. sorry

 

[Justin Lazear]

$$x - \left(\ln{x^2}\right) - 1$$
or
$$x - \ln{(x^2 -1)}$$
?

Well, either way, the limit is still unbounded.

--J

 

[mad]

$$x - (\ln{x^2 -1})$$
?

This one. I will try your method. Is it the only way?

 

[Justin Lazear]

Heh, I screwed up the notation, too. I edited to give the proper two possibilities. As it was before, they said exactly the same thing!

--J