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?

A very standard trick for calculating limits when L'Hopital or Taylor series are not allowed is using the definition of the derivative of some function at some point. Usually, this "some point" is the point at which you take the limit and the "some function" is the "complicated" one in the denominator. 
In your case, if you look at the definition of the derivative of the function $f(x)=\log_2(x)$ at $x=2$, you have
$$
f'(2)=\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=
\lim_{x\to 2}\frac{\log_2x-1}{x-2}.\tag{1}
$$
Note that (1) is not far away from the limit you want! Acutally, you have
$$
\frac{\log_2x-1}{x^2-4}=\frac{\log_2x-1}{x-2}\cdot\frac{1}{x+2}.\tag{2}
$$
On the other hand, 
$$
f'(2)=\frac{1}{x\ln 2}\bigg|_{x=2}=\frac{1}{2\ln 2}.\tag{3}
$$
Now you can combine (1) (2) and (3) to get your answer. (Note that in your answer, $\ln 256=8\ln 2$.)
