\(\mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{{x^2} + x - 2}}{{x - 2}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{{x^2} + x - 2}}{{x - 2}}\).
Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} + x - 2} \right) = {2^2} + 2 - 2 = 4 > 0\\\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 2 - 2 = 0\\x > 2 \Rightarrow x - 2 > 0\end{array} \right.\)
Vậy \(\mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{{x^2} + x - 2}}{{x - 2}} = + \infty \).
Chọn C.