# Thread: Quick limit question

1. ## Quick limit question

So I've boiled down a problem to the point that I've got ln|x-1|-ln|x+1| evaluated at B=-2 and A=-infinity

Boiling it down further, I get ln|(-3)| - ln|-1|-(limits as X->-inf for both of the terms)

This gets me to LN(3)-0-(infinity-infinity). Is that valid? It says the answer is LN(3) but can I really subtract infinity from infinity and get zero?

2. Originally Posted by Wolvenmoon
So I've boiled down a problem to the point that I've got ln|x-1|-ln|x+1| evaluated at B=-2 and A=-infinity

Boiling it down further, I get ln|(-3)| - ln|-1|-(limits as X->-inf for both of the terms)

This gets me to LN(3)-0-(infinity-infinity). Is that valid? It says the answer is LN(3) but can I really subtract infinity from infinity and get zero?
Post the original question.

CB

3. Original question is the integral from -infinity to -2 of 2dx/(x^2-1)

4. Originally Posted by Wolvenmoon
Original question is the integral from -infinity to -2 of 2dx/(x^2-1)
$\displaystyle \ln|x-1|-\ln|x+1|=\ln\left|\frac{x-1}{x+1}\right|$
As $\displaystyle x\to -\infty$ that limit is $\displaystyle 0$.
So your answer is just $\displaystyle \ln(3)$.