# Math Help - Limit of the ln of an absolute value

1. ## Limit of the ln of an absolute value

lim ln|x -2|
x--> 2

I'm not quite sure what to do. I know that I can't take the ln of a negative number and I can't take the ln of 0. When I evaluate the limit for x = 2, my function is undefined and so is my limit (did I say this correctly?).

Am I correct?

Even if I manipulate the function knowing the definition of absolute value i.e. |x| = x^2/2 i still get

ln(x-2)^(2/2)
ln(x-2)^(1/2) + ln(x-2)^(1/2)

2. Originally Posted by Noxide
lim ln|x -2|
x--> 2

I'm not quite sure what to do. I know that I can't take the ln of a negative number and I can't take the ln of 0. When I evaluate the limit for x = 2, my function is undefined and so is my limit (did I say this correctly?).

Am I correct?

Even if I manipulate the function knowing the definition of absolute value i.e. |x| = x^2/2 i still get

ln(x-2)^(2/2)
ln(x-2)^(1/2) + ln(x-2)^(1/2)
you are not at x = 2, this is a limit, you are getting arbitrarily close to x = 2. the absolute value keeps everything positive, so the argument of the log is never negative. this limit is essentially the same as $\lim_{t \to 0^+} \ln t = - \infty$ (look at the graph)

and that's not the definition of absolute value: $|x| = \sqrt {x^2} = \left \{ \begin{array}{ll} x & \text{ if } x \ge 0 \\ & \\ -x & \text{ if } x < 0 \end{array} \right.$

3. I think you should look at the limit from another perspective.

remember the definition of a limit (i don't think i can write it out here, its the one with epsilon and delta)

the |x-2| as x-->2 approaches 0 from both sides, and as ln(x) x-->0, it approaches negative infinity. and since this is for both sides, the limit exists at neg. inf.

4. Doesn't the limit have to be a finite element of the reals though?

5. Originally Posted by Noxide
Doesn't the limit have to be a finite element of the reals though?
no. there is such a thing as infinite limits. it should be defined in your text somewhere. your professor may prefer you to answer the problem with "no limit", but that is between you and your professor. it's kind of like when you were in beginning algebra and they told you that if you encountered $x^2 + 1 = 0$ write "no solution" (what they really meant was there was no real solution, we can actually solve this, $x = \pm i$)