$\large \lim\limits_{x\to 2}\frac{x-2}{2 + \log_2(3) - x -\log_2(2x-1)} $ determination 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:


*

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

*dividing all stuff by (x-2)
$\lim\limits_{x\to2}\dfrac{1}{-1 + \dfrac{\log_2(3) - \log_2(2x-1)}{x-2}} $

*change variable $x-2=u$ so $u\to0$
$\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{\log_2(3) - \log_2(2u+3)}{u}} $

*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.

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