Natural Domains and Limits

**Pandora** · August 11th 2008, 11:52 AM

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
Thanks in advance

**earboth** · August 11th 2008, 12:28 PM

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)?

$f(x)=\ln(\ln(x))$ then

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

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

Therefore

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

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

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

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

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

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

**Pandora** · August 11th 2008, 05:31 PM

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.

**earboth** · August 11th 2008, 09:45 PM

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:

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

Now it is obvious that

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

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

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

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