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Math Help - limit of ln

  1. #1
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    limit of ln

    What is the limit as x approaches a for, (ln(x)-ln(a))/(x-a)
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: limit of ln

    \lim_{x\to a} \, \frac{\log (x)-\log (a)}{x-a}=1
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  3. #3
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    Re: limit of ln

    Isn't that the definition of the derivative of \log{x} at x=a? So it would be 1/a.
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  4. #4
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    Re: limit of ln

    Quote Originally Posted by calman View Post
    What is the limit as x approaches a for, (ln(x)-ln(a))/(x-a)
    As Hollywood posted, they want you to see that it's the derivative of f(x) = ln(x) at x = a, and so is 1/a. I've usually seen that done by defining the natural log as an integral, and then taking its derivative using the Fundamental Theorem of Calculus.

    However, if you only know about the derivative of e^x, and have ln(x) defined as its inverse, you could do this:

    Let f(x) = e^x. Since f^{-1}(f(x)) = x \ \forall x \in \mathbb{R}, take derivatives to get (f^{-1})'(f(x))(f'(x)) = 1 \ \forall x \in \mathbb{R}.

    Since f(x) = e^x, that reads (f^{-1})'(f(x)) = 1/f(x) \ \forall x \in \mathbb{R}, so

    (f^{-1})'(e^x) = 1/(e^x) \ \forall x \in \mathbb{R}, so, since the image of f is all positive reals, have

    (f^{-1})'(t) = 1/t \ \forall t> 0, t \in \mathbb{R}.

    Also, could do this:

    \lim_{x \to a} \frac{ln(x)-ln(a)}{x-a} = \lim_{x \to a} \frac{ln(x/a)}{a\left(\frac{x}{a}-1\right)}

    = \frac{1}{a} \ \lim_{x \to a} \frac{ln(x/a)}{\frac{x}{a}-1} = \frac{1}{a} \ \lim_{t \to 1} \frac{ln(t)}{t-1}

    =(substituting t = e^s) \frac{1}{a} \ \lim_{s \to 0} \frac{ln(e^s)}{e^s-1} = \frac{1}{a} \ \lim_{s \to 0} \frac{s}{e^s-1}

    = \frac{1}{a} \ \lim_{s \to 0} \frac{1}{\frac{e^s-1}{s}} = \frac{1}{a} \ \frac{1}{\lim_{s \to 0} \frac{e^s-1}{s}}

    = \frac{1}{a} \ \frac{1}{1} = \frac{1}{a}
    Last edited by johnsomeone; October 8th 2012 at 12:51 PM.
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  5. #5
    Senior Member MaxJasper's Avatar
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    Re: limit of ln

    Sorry folks, /a was dropped during Latex editing!
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