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Thread: Limits - Explanation of a step.

  1. #1
    Super Member Deadstar's Avatar
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    Limits - Explanation of a step.

    Perhaps someone could explain the last step here for me please...

    Find the limit of $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x)}{x-1}$. Do not use L'Hopital's rule.

    Solution;
    Since In(1) = 0, we have that,

    $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x)}{x-1}$ = $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x) - In(1)}{x-1}$ = $\displaystyle \frac{d}{dx} In(x) = 1$ for x =1. (Not quite sure how to do the latex for the way its written in my notes. To the right of the d/dx is a vertical line with x=1 at the bottom.)

    And... Im not quite sure how to do the last step there i.e the $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x) - In(1)}{x-1}$ = $\displaystyle \frac{d}{dx} In(x) = 1$ bit...

    Any help on this?
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  2. #2
    o_O
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    First off, the natural logarithm is abbreviated $\displaystyle \ln (x)$ (an L not an I).

    Recall the definition of a derivative of a function at some point x = a:
    $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

    So imagine: $\displaystyle f(x) = \ln (x)$ and a = 1.
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  3. #3
    Super Member Deadstar's Avatar
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    jeez thats so simple..! Cheers for that.
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  4. #4
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    We start with a list of substitutions:

    a) Let $\displaystyle u=x-1,$ and the limit becomes $\displaystyle \underset{u\to 0}{\mathop{\lim }}\,\frac{\ln (1+u)}{u}=\underset{u\to 0}{\mathop{\lim }}\,\ln (1+u)^{1/u}.$

    b) Let $\displaystyle z=\frac1u,$ and the limit becomes $\displaystyle \underset{z\to \infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{z} \right)^{z}=\ln e=1.$
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