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The limit of $\lim\limits_{x\to2}\dfrac{x-2}{\;2 + \log_2(3) - x - \log_2(2x-1)} $ is $\dfrac{-4}{\ln (2e)}$ i need to show that. I had try this:

  1. rearrange the denominator

    $\lim\limits_{x\to2}\dfrac{x-2}{-(x-2) + \log_2(3) - \log_2(2x-1)} $

  2. dividing all stuff by (x-2)

    $\lim\limits_{x\to2}\dfrac{1}{-1 + \dfrac{\log_2(3) - \log_2(2x-1)}{x-2}} $

  3. change variable $x-2=u$ so $u\to0$

    $\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{\log_2(3) - \log_2(2u+3)}{u}} $

  4. simplify

    $\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{- \log_2(2/3u+1)}{u}} $

5 changing the log base to e

$\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{- \ln(2/3u+1)}{2/3u}\times \dfrac{2}{3 \ln 2}} $

6.the final solution is $\dfrac{1}{-1 - 1\times \dfrac{2}{3 \ln 2}} = (-3/2) \ln2 $

this is not the correct answer. Why? where is my mismatch.

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It's just a small error at the end: $$ \frac{1}{-1 - 1 \left( \frac{2}{3 \ln 2} \right)} = \frac{-3 \ln 2}{3 \ln 2 + 2} \ne -\frac{3}{2} \ln 2 $$

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  • $\begingroup$ Thanks, but the correct answer is $\dfrac{-4}{ln( 2e)}$ $\endgroup$ – Jorge Feb 7 '14 at 23:55
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    $\begingroup$ No, it's not. The limit of the function you wrote is $-3 \log 2 / (3 \log 2 + 2)$, as correctly stated by Goos. $\endgroup$ – heropup Feb 7 '14 at 23:56
  • $\begingroup$ I'm happy for that. the book solution is wrong. I was going crazy with that limit. $\endgroup$ – Jorge Feb 8 '14 at 0:01
  • $\begingroup$ For extra credit, figure out what the anonymous solver for the textbook did wrong... ;) [unless this was another case of replacing a problem and forgetting to replace the answer in the back...] $\endgroup$ – colormegone Feb 8 '14 at 0:33
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The limit is much easier to calculate if we recall the definition of derivative: $$f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}.$$ Then with $a = 2$, and $f(x) = \log_2 (2x-1),$ we obtain $$\lim_{x \to 2} \frac{\log_2(2x-1) - \log_2 3}{x-2} = \frac{d}{dx}\left[\log_2 (2x-1) \right]_{x=2} = \left[\frac{2}{(2x-1)\log 2}\right]_{x=2} = \frac{2}{3 \log 2}.$$ It follows that $$\begin{align*} \lim_{x \to 2} \frac{x-2}{(2-x) + \log_2 3 - \log_2 (2x-1)} &= -(1 + f'(2))^{-1} \\ &= -\left( 1 + \frac{2}{3 \log 2} \right)^{\!-1} \\ &= - \frac{3 \log 2}{2 + 3 \log 2}. \end{align*}$$

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  • $\begingroup$ thanks for this alternative approach. $\endgroup$ – Jorge Feb 8 '14 at 0:04
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There is another way using replacing $\log (2 x-1)$ by its Taylor series built around $x=2$. Limited to the first term, this expansion is $$\log (2 x-1)=\frac{2 (x-2)}{3}+\log (3)$$ The original expression being rewritten as $$\frac{(x-2)}{(2-x)-\frac{\log (2 x-1)}{\log (2)}+\frac{\log (3)}{\log (2)}}$$So, replacing the logarithmic term in the denominator leads to an expression from which all $(x-2)$ terms can be removed and what is left is the limit $$-\frac{\log (2)}{\frac{2}{3}+\log (2)}$$ which is identical to what other answers already gave.

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