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Math Help - Proving!

  1. #1
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    Proving!

    Hi I have two questions here, which I do not know how to prove.

    1.) Prove that (lnx)/x ≤ 1/e for all x>0.

    I know to find the maximum value of (lnx)/x, I got x=e. But then what do I do? I subbed x=e back into (lnx)/x and I got 1/e.. 1/e ≤ 1/e doesn't make sense..

    2.) Prove that lnx ≤ x-1 for all x>0.
    Do I find the maximum value for lnx? I can't really get the maximum value for I'll get 1 = 0.

    Please help, thank you so so much!
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  2. #2
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    Re: Proving!

    Question 2 just requires you to know that \displaystyle \begin{align*} \lim_{x \to 0^{+}} \ln{(x)} = -\infty \end{align*} while \displaystyle \begin{align*} \lim_{x \to 0^{+}} (x - 1) = -1 \end{align*}, which means \displaystyle \begin{align*} \ln{(x)} \end{align*} starts out less than \displaystyle \begin{align*} x - 1 \end{align*}, and also that \displaystyle \begin{align*} \ln{(x)} \end{align*} grows slower than \displaystyle \begin{align*} x-1 \end{align*}.
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  3. #3
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    Re: Proving!

    Quote Originally Posted by Tutu View Post
    1/e ≤ 1/e doesn't make sense.
    Why does is not make sense?
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  4. #4
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    Re: Proving!

    If the maximum value of the function is 1/e then obviously for all x the function value is < 1/e.
    You cannot find the maximum value for ln x. It is a monotonically increasing function that asymptotically goes to \infinity.
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  5. #5
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    Re: Proving!

    Quote Originally Posted by Tutu View Post
    2.) Prove that lnx ≤ x-1 for all x>0.
    There is a famous inequality.
    Use the mean value theorem to prove that.
    If 0<a<b then \frac{1}{b}\le\frac{\ln(b)-\ln(a)}{b-a}\le\frac{1}{a}.
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