The limit of the function $ \frac{\log_{2}(x)-1}{x^2-4}$ as $x=2$ I want to solve the limit:
$$\lim_{x \to 2}{\frac{\log_{2}(x)-1}{x^2-4}} $$
The problem is I can't use L'Hospital and Taylor Series to solve it.
In My attempt I'm stuck here:
$$ \lim_{x \to 2}{\frac{\frac{\ln(x)}{\ln(2)}-1}{x^2-4}} $$
$$ \lim_{x \to 2}{\frac{\frac{-\ln(2)+\ln(x)}{\ln(2)}}{(x-2)(x+2)}} $$
I know the answer is: $\frac{1}{\ln(256)}$ , but how can I make a substitution to get there?

HINT
We have that
$$ \lim_{x \to 2}{\frac{\log_{2}(x)-1}{x^2-4}}=\lim_{x \to 2}\frac1{x+2}{\frac{\log_{2}(x)-\log_2 2}{x-2}}$$
then apply the definition of derivative or, as an alternative, by $x=y+2$ with $y\to 0$ we have
$$\lim_{x \to 2}{\frac{\log_{2}(x)-\log_2 2}{x-2}}=\lim_{y \to 0}{\frac{\log_{2}(1+y/2)}{y}}=\lim_{y \to 0}{\frac1{2\log 2}\frac{\log (1+y/2)}{y/2}}$$
