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Find the limit as x goes to 1 of (1 / log x – 1 / (x-1))

Evaluate the limit.

    \[ \lim_{x \to 1} \left( \frac{1}{\log x} - \frac{1}{x-1} \right). \]

We use this exercise (Section 7.11, Exercise #4) to evaluate,

    \begin{align*}  \lim_{x \to 1} \left( \frac{1}{\log x} - \frac{1}{x-1} \right) &= \lim_{x \to 1} \left( \frac{x - 1 - \log x}{(x-1) \log x} \right) \\[9pt]  &= \lim_{x \to 1} \left( \frac{(x-1) - (x-1) + \frac{1}{2}(x-1)^2 + o((x-1)^2)}{(x-1)\left((x-1) - \frac{1}{2}(x-1)^2 + o((x-1)^2) \right)}\right) \\[9pt]  &= \lim_{x \to 1} \left( \frac{\frac{1}{2}(x-1) + o((x-1)^2)}{(x-1) - \frac{1}{2}(x-1)^2 + o((x-1)^2)} \right) \\  &= \lim_{x \to 1} \left( \frac{\frac{1}{2} + o((x-1)^2)}{1 - (x-1) + o((x-1)^2)} \right) \\[9pt]  &= \frac{1}{2}. \end{align*}


  1. Marko says:

    I have worked through exercises nr. 6 – 29 and It actually isn’t that complicated. I have one thing however that confuses me and caused some problems (especially at ex. nr. 20 -22). How do I know to what degree I have to take the polynomial? The choices in the solutions sometimes seemed arbitrary to me. I couldn’t see a pattern there. Because of that I sometimes chose to many degrees which resulted in an algebraic mess and sometimes to few degrees which when I evaluated the expression led me back to 0/0 or similiar. Is that something that comes with experience or is there a fixed set or rules that tells me to what degree I have to take the polynomial in each case?

    • Artem says:

      This clearly is trial and error until you get the right expansion. If you do 100 of these, maybe you can easily see the required expansion then.

      • Anonymous says:

        Second step I am not getting. I can’t understand that step… can u plz give more explanation for that 2nd step?

      • Vaishnavi says:

        I can’t understand 2nd step from that solution…. can u plz give more explanation about that 2nd step?

      • S says:

        It is probably simpler if you just substitythe variable x=t+1 and take the limit approaching 0.

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