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Math Help - Natural Domains and Limits

  1. #1
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    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
    Thanks in advance
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  2. #2
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    Quote Originally Posted by Pandora View Post
    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.
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  3. #3
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    .

    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.
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  4. #4
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    Quote Originally Posted by Pandora View Post
    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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