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Math Help - l'Hopital's Rule

  1. #1
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    l'Hopital's Rule

    I'm supposed to use l'Hopital's rule to calculate:
    lim x-->infinity: [ln(ln(x))]/[xln(x)]

    I know I find the derivative of the numerator and the derivative of the denominator.

    For the numerator: Dx ln(ln(x)) = 1/(ln(x))
    For the denominator (product rule since I have two terms, x, and ln(x) ) Dx xln(x) = (ln(x)) (x(1/x))

    I'm not sure if that's correct or not, but I know I have to actually calculate the limit and that doesn't seem possible with the way the equations stands after the first derivative of the numerator and denominator. If somebody could show me what I have to do next that would be great.

    Thanks
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  2. #2
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    Quote Originally Posted by Adrian View Post
    I'm supposed to use l'Hopital's rule to calculate:
    lim x-->infinity: [ln(ln(x))]/[xln(x)]

    I know I find the derivative of the numerator and the derivative of the denominator.

    For the numerator: Dx ln(ln(x)) = 1/(ln(x))
    For the denominator (product rule since I have two terms, x, and ln(x) ) Dx xln(x) = (ln(x)) (x(1/x))

    I'm not sure if that's correct or not, but I know I have to actually calculate the limit and that doesn't seem possible with the way the equations stands after the first derivative of the numerator and denominator. If somebody could show me what I have to do next that would be great.

    Thanks
    \lim_{x\to \infty} \frac{\ln (\ln x)}{x\ln x}  = \lim_{y\to \infty} \frac{\ln y}{e^y y}=0.
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  3. #3
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    Differentiating the num, one gets \frac{1}{xln(x)}

    The den, one gets ln(x)+1

    Then we have \lim_{x\rightarrow{\infty}}\frac{1}{xln(x)(ln(x)+1  )}

    Now, see the limit?.
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  4. #4
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    Quote Originally Posted by galactus View Post
    Differentiating the num, one gets \frac{1}{xln(x)}

    The den, one gets ln(x)+1

    Then we have \lim_{x\rightarrow{\infty}}\frac{1}{xln(x)(ln(x)+1  )}

    Now, see the limit?.
    I suppose that the denominator is going to infinty meaning the limit will be 0. However, could you explain how you came to your solutions for the numerator and denominator, I tried to figure them out, but I couldn't see how you worked them out.

    Thanks again for all the help.
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  5. #5
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    I just differentiated the top and bottom.

    Derivative of ln(ln(x)) using chain rule.

    The derivative of ln(x) is 1/x, so with the chain rule we have \frac{1}{xln(x)}

    Using product rule, the derivative of xln(x) is x(\frac{1}{x})+ln(x)=ln(x)+1

    Dividing, we get \frac{1}{xln(x)(ln(x)+1)}

    So, you are correct, the limit is 0.
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  6. #6
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    Here is another non L'Hopital way:
    \left| \frac{\ln (\ln x)}{x\ln x} \right| \leq \frac{\ln x}{x\ln x} = \frac{1}{x} \to 0
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  7. #7
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    L'Hopital is a nice fall back, but I am like PH on this. I like to find them without it if possible. And it's mostly possible. Since you're required to learn L'Hopital, go with it.
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  8. #8
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    Right, I forgot about the chain rule, thanks a bunch guys.
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  9. #9
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    While we're on the topic of l'Hopital, I saw a problem in my textbook (with no solution) that looked like this:

    lim x--> infinity: (3x-4)/(2x-5)

    If I differentiate the numerator and denominator I get 3/2 for the limit as x goes to infinity, would this be correct? It seems kind of simple, but then again, maybe it was an introductory problem to the concept of l'Hopital's rule.

    Thanks once again.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Adrian View Post
    While we're on the topic of l'Hopital, I saw a problem in my textbook (with no solution) that looked like this:

    lim x--> infinity: (3x-4)/(2x-5)

    If I differentiate the numerator and denominator I get 3/2 for the limit as x goes to infinity, would this be correct? It seems kind of simple, but then again, maybe it was an introductory problem to the concept of l'Hopital's rule.

    Thanks once again.
    yes, the limit is 3/2, and there are several ways to get this without L'Hopital's
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  11. #11
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    Yes, that's correct. You certaily don't need the 'Hospital' rule.

    Possibly, an easy, early one to introduce...as you said.
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