Đáp án đúng: A

Phương pháp giải:
Giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) tồn tại khi và chỉ khi \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\).
Giải chi tiết:Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{\sin \left( {x - 2} \right)}}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{\sin \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{1}{{x - 1}}\dfrac{{\sin \left( {x - 2} \right)}}{{x - 2}} = \dfrac{1}{{2 - 1}}.1 = 1\\\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{\sqrt {2{x^2} + 8}  - x - 2}}{{{x^2} - 4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{\sqrt {2{x^2} + 8}  - 4 - x + 2}}{{{x^2} - 4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{\sqrt {2{x^2} + 8}  - 4}}{{{x^2} - 4}} - \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{x - 2}}{{{x^2} - 4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{2{x^2} + 8 - 16}}{{\left( {{x^2} - 4} \right)\left( {\sqrt {2{x^2} + 8}  + 4} \right)}} - \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{1}{{x + 2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{2{x^2} - 8}}{{\left( {{x^2} - 4} \right)\left( {\sqrt {2{x^2} + 8}  + 4} \right)}} - \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{1}{{x + 2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{2}{{\sqrt {2{x^2} + 8}  + 4}} - \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{1}{{x + 2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{{\sqrt {{{2.2}^2} + 8}  + 4}} - \dfrac{1}{{2 + 2}} = 0\end{array}\)
Vì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)
e \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\) nên không tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).
Vậy \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 1,\,\,\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 0\) và không tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).
Chọn A.