# Thread: Natural Domains and Limits

1. ## Natural Domains and Limits

Hi.
I'm pretty new to this thread, but am hoping that you guys can help me out.
Would the natural domain of 1/( 1-x) function be all real numbers apart from 1? This whole concept of natural domain seems to be simple, and yet this question is a bit fuzzy. Also, Would I be right in saying that the natural domain of f= ln(lnx) be 1, infinity? If that is the case, what would be the natural domain of f o f? And how would one go about finding the axes intercepts of y=(f o f)(x)?

Hope it is not too much trouble

2. Originally Posted by Pandora
Hi.
I'm pretty new to this thread, but am hoping that you guys can help me out.
Would the natural domain of 1/( 1-x) function be all real numbers apart from 1? Correct
... Would I be right in saying that the natural domain of f= ln(lnx) be 1, infinity? Yes
If that is the case, what would be the natural domain of f o f? And how would one go about finding the axes intercepts of y=(f o f)(x)?
$\displaystyle f(x)=\ln(\ln(x))$ then

$\displaystyle y = (f\circ f)(x) = \ln(\ln(\ln(\ln(x))))$

To calculate the domain you have to keep in mind that the ln-function is defined for all positive real numbers: $\displaystyle d = (0, \infty)$

Therefore

$\displaystyle \ln(\ln(\ln(\ln(x)))) > 0$ that means

$\displaystyle \ln(\ln(\ln(x))) > 1$ that means

$\displaystyle \ln(\ln(x)) > e$ that means

$\displaystyle \ln(x) > e^e$ that means

$\displaystyle x> e^{e^e}$ that means $\displaystyle x > \approx 3,814,279.105$

That is really an unexpected result. So please check my considerations and my arithmetic.

3. ## .

Thank you, I get it now. I think I will leave it at e^e^e.

I already have another question in store though.. for the function, (90e^x)/(e^x+5), i know that it will not increase without bound because I drew it on a graphics calculator. However, how do I prove it using limits?

Oh, and does the function have horizontal or vertical asymptotes? I feel way out of my depth here.

4. Originally Posted by Pandora
Thank you, I get it now. I think I will leave it at e^e^e.

I already have another question in store though.. for the function, (90e^x)/(e^x+5), i know that it will not increase without bound because I drew it on a graphics calculator. However, how do I prove it using limits?

Oh, and does the function have horizontal or vertical asymptotes? I feel way out of my depth here.
Re-write the term of the function:

$\displaystyle 90e^x \div (e^x+5) = 90 - \dfrac{450}{e^x+5}$

Now it is obvious that

$\displaystyle \lim_{x \rightarrow \infty} \left(90 - \dfrac{450}{e^x+5} \right)= 90$ because

$\displaystyle \lim_{x \rightarrow \infty} \left( \dfrac{450}{e^x+5} \right)= 0$

Since $\displaystyle e^x+5 > 5 \neq 0$ the function doesn't have any vertical asymptotes. By the calculation of the limit you see that there exist two horizontal asymptotes:

1. y = 90

2. $\displaystyle \lim_{x \rightarrow -\infty} \left( \dfrac{90e^x}{e^x+5} \right)= 0 ~\implies~ \boxed{y = 0}$