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Math Help - why does the limit of ln(x)/x approach 0???

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    why does the limit of ln(x)/x approach 0???

    the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1
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    \lim_{x \to \infty}\frac{\ln{x}}{x}.

    Since this goes to \frac{\infty}{\infty} you can use L'Hospital's Rule.


    \lim_{x \to \infty}\frac{\ln{x}}{x} = \lim_{x \to \infty}\frac{\frac{d}{dx}(\ln{x})}{\frac{d}{dx}(x)  }

     = \lim_{x \to \infty}\frac{\frac{1}{x}}{1}

     = \lim_{x \to \infty}\frac{1}{x}

     = 0.
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    oh DUHHHH because infinite/infinite DOES NOT equal 1; it is indeterminate!!!
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    Quote Originally Posted by RedSwiss View Post
    the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1
    get that idea out of your head.
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  5. #5
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    Quote Originally Posted by RedSwiss View Post
    the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1
    Even though both numerator and denominator are growing without bound, the real question is which is growing faster.

    Also note that for x \ge e

    1 \le \ln x \le \sqrt{x}

    so

    \dfrac{1}{x} \le  \dfrac{\ln x}{x} \le    \dfrac{1}{\sqrt{x}},

    Thus,

    \displaystyle \lim_{x \to \infty} \dfrac{1}{x} \le   \lim_{x \to \infty}  \dfrac{\ln x}{x} \le   \lim_{x \to \infty}  \dfrac{1}{\sqrt{x}}

    which gives

    \displaystyle   \lim_{x \to \infty}  \dfrac{\ln x}{x}} = 0.

    A non L'Hopital way of doing it.
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    Quote Originally Posted by RedSwiss View Post
    oh DUHHHH because infinite/infinite DOES NOT equal 1; it is indeterminate!!!
    Well, I'm glad you said that. It would have sounded harsh if I had said it!

    Note that ln(x) and x both go to infinity as x goes to infinity. But x goes to infinity faster than ln(x) does.
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