$\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.

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 $$\begin{align*} \lim_{x \to 2} \frac{x-2}{(2-x) + \log_2 3 - \log_2 (2x-1)} &= -(1 + f'(2))^{-1} \\ &= -\left( 1 + \frac{2}{3 \log 2} \right)^{\!-1} \\ &= - \frac{3 \log 2}{2 + 3 \log 2}. \end{align*}$$
