Hello.

I do not understand how:

$ln|\frac{\infty-2}{\infty+2}|$ is equal to $ln(1)$?

Does $ln(\infty)$ always = $ln(1)$?

Thank you,
Jen

I am so thankful for MathMan's post on how to use LaTeX!

*I apologize...I forgot to include the context I'm asking this for...

$\int\frac{1}{z^2-4}dz, (10

= $\frac{1}{4}ln|\frac{z-2}{z+2}|$ with 10<z< $\infty$

So,

$\frac{1}{4}ln|\frac{\infty-2}{\infty+2}|-\frac{1}{4}ln(\frac{2}{3})$

should equal:

$-\frac{1}{4}ln(\frac{2}{3}$): converges to .1014

2. Originally Posted by Jenberl
Hello.

I do not understand how:

$ln|\frac{\infty-2}{\infty+2}|$ is equal to $ln(1)$?

Does $ln(\infty)$ always = $ln(1)$?

Thank you,
Jen

I am so thankful for MathMan's post on how to use LaTeX!

*I apologize...I forgot to include the context I'm asking this for...

$\int\frac{1}{z^2-4}dz, (10

= $\frac{1}{4}ln|\frac{z-2}{z+2}|$ with 10<z< $\infty$

So,

$\frac{1}{4}ln|\frac{\infty-2}{\infty+2}|-\frac{1}{4}ln(\frac{2}{3})$

should equal:

$-\frac{1}{4}ln(\frac{2}{3}$): converges to .1014

OK, first off, you can NOT treat infinity as a number.

The correct notation is $\lim_{\varepsilon \to \infty} \frac{1}{4}\ln{\left |\frac{\varepsilon - 2}{\varepsilon + 2}\right | }$.

Notice that $\frac{z - 2}{z + 2} = 1 - \frac{4}{z + 2}$.

What happens as $z \to \infty$? The denominator gets huge, so the fraction $\to 0$.

So it should tend to $\ln{1} = 0$.

3. Originally Posted by Prove It
OK, first off, you can NOT treat infinity as a number.

The correct notation is $\lim_{\varepsilon \to \infty} \frac{1}{4}\ln{\left |\frac{\varepsilon - 2}{\varepsilon + 2}\right | }$.

Notice that $\frac{z - 2}{z + 2} = 1 - \frac{4}{z + 2}$.

What happens as $z \to \infty$? The denominator gets huge, so the fraction $\to 0$.

So it should tend to $\ln{1} = 0$.

OH!
That makes a lot more sense now.
Also- thank you for showing (inadvertantly) how to express absolute value notation properly with '\left' & '\right' before the '|' .

~Jen