Answer
(a)$$\lim_{x \to 2^-}\frac{x^2-4}{|x-2|}=-4$$
(b)$$\lim_{x \to 2^+}\frac{x^2-4}{|x-2|}=4$$
(c)
$\lim_{x \to 2}\frac{x^2-4}{|x-2|}$ does not exist.
Work Step by Step
(a)
For $x<2$, we have $x-2<0$ and so $|x-2|=-(x-2)$. Thus, we get$$\lim_{x \to 2^-}\frac{x^2-4}{|x-2|}=\lim_{x \to 2^-}\frac{(x+2)(x-2)}{-(x-2)}=\lim_{x \to 2^-}-(x+2)=-4.$$
(b)
For $x>2$, we have $x-2>0$ and so $|x-2|=x-2$. Thus, we get$$\lim_{x \to 2^+}\frac{x^2-4}{|x-2|}=\lim_{x \to 2^+}\frac{(x+2)(x-2)}{x-2}=\lim_{x \to 2^+}(x+2)=4.$$
(c)
Since $\lim_{x \to 2^-}\frac{x^2-4}{|x-2|} \neq \lim_{x \to 2^+} \frac{x^2-4}{|x-2|}$, $\lim_{x \to 2}\frac{x^2-4}{|x-2|}$ does not exist.
