finding the limit right answer wrong sign

Question

I have the following equation

Given $$\lim_{x\to 2}\frac{2-x}{x^2-4}$$

using substitution we know that both the top and the bottom solve to $\frac{0}{0}$ this means that (per my text book and this is where I am lost) that the denominator is a factor of the numerator (it seems the numerator is a factor here...).

The above can be rewritten as

$$\frac{x-2}{(2-x)(x+2)}$$

cancel out the common factors

$$\frac{1}{x+2}$$

which solves to $\frac{1}{4}$

which is totally not right, as my answer key tells me that it is $-\frac{1}{4}$

Somewhere along the lines I am missing a key step.

Answer 1

In the comments, you said the question actually was

$$\lim_{x\to 2}\frac{2-x}{x^2-4}$$

Which since $2-x=-(x-2)$ rewrites to

$$-\lim_{x\to 2}\frac{x-2}{x^2-4}$$

This limit you already found as $1/4$ so

$$\lim_{x\to 2}\frac{2-x}{x^2-4}=-\frac14$$

Answer 2

$$\lim_{x\to 2}\frac{2-x}{x^2-4}=\lim_{x\to 2}\frac{2-x}{(x-2)(x+2)}=\lim_{x\to 2}\frac{-1}{x+2}=-\frac{1}{4}$$

OR

Using L'Hospital rule, $$\lim_{x\to 2}\frac{2-x}{x^2-4}=\lim_{x\to 2}{\Large\{}\frac{\frac{d}{dx}2-x}{\frac{d}{dx}x^2-4}{\Large\}}=\lim_{x\to 2}\frac{-1}{2x}=-\frac{1}{4}.$$