1. ## lost of issues

I meant lots of issues, sorry

The problem says to : Evaluate the improper integrals. Show the proper use of limits.

The integral of 1/(x^2-9) from 3 to infinity.

I can get it into the limit form and solve the integral part but I can't solve the limit. It works out to the limit as a approaches 3 from the left(+) of (1/9 ln 1/4 - 1/9 ln |(a-3)/(a+3)|) + limit as b approaches infinity from the right (-) of (1/9 ln |(b-3)/(b+3)| - 1/9 ln 1/4) How do you solve that?

Also I can't figure out that the limit of e^(-1/x) as x approaches 0 from the right and from the left is.

Finially I don't know where to even start on the integral of
sqrt(9-(lnx)^2)
(lnx^x)

2. $\displaystyle \int_{3}^{\infty} \frac{1}{(x^2-9)} = -1/3 \tanh^{-1} (x/3).$

$\displaystyle \lim_{x\rightarrow \infty} -1/3 \tanh^{-1} (x/3) + \lim_{x\rightarrow 3}1/3 \tanh^{-1} (x/3)$

Now just use the definition $\displaystyle \tanh^{-1}(x) = 1/2[\ln(1+x) - \ln(1-x)].$

$\displaystyle \lim_{x \rightarrow 0} e^{(-1/x)} = \lim_{x \rightarrow 0} \frac{1}{e^{(1/x)}} = \frac{1}{\infty} = 0$

3. Originally Posted by Anonymous1
$\displaystyle \int_{2}^{\infty} \frac{1}{(x^2-9)} = -1/3 tanh^{-1} (x/3).$

I assume you want $\displaystyle \lim_{x\rightarrow +-3} -1/3 tanh^{-1} (x/3)?$
I typed it wrong the lower number should have been 3 not 2. I'm sorry.

4. I solved the integral part to be 1/9 ln |(x-3)/(x+3)|

5. So what are you taking the limit of?

6. Originally Posted by Anonymous1
So what are you taking the limit of?
I hope this makes sense. I don't know how to type it.

∫ 1/(x^2-9) from 3 to infinity, ok so the 3 and the infinity make it improper, so it becomes:
limit as a(bottom number) approaches 3 of ∫ 1/(x^2-9) w/ the bottom number being 3 and the top number being any number you choose, so let's say 5 from the left(+). + limit as b(top number approaches infinity of ∫ 1/(x^2-4) with the top number being infinity and the bottom number being the 5 I picked from the right (-).

So it is the limit as a approaches 3 from the left(+) of (1/9 ln |(5-3)/(5+3)| - 1/9 ln |(a-3)/(a+3)|) + limit as b approaches infinity from the right (-) of (1/9 ln |(b-3)/(b+3)| - 1/9 ln |(5-3)/(5+3)|), which is

the limit as a approaches 3 from the left(+) of (1/9 ln 1/4 - 1/9 ln |(a-3)/(a+3)|) + limit as b approaches infinity from the right (-) of (1/9 ln |(b-3)/(b+3)| - 1/9 ln 1/4)

After that I'm lost