\(\mathop {\lim }\limits_{x \to 5} \dfrac{{x + 3}}{{3 - x}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 5} \dfrac{{x + 3}}{{3 - x}}\)
Ta có:
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 5} \left( {\dfrac{{x + 3}}{{3 - x}} + 4} \right) = \mathop {\lim }\limits_{x \to 5} \left( {\dfrac{{x + 3 + 12 - 4x}}{{3 - x}}} \right)\\ = \mathop {\lim }\limits_{x \to 5} \dfrac{{ - 3x + 15}}{{3 - x}} = \mathop {\lim }\limits_{x \to 5} \dfrac{{ - 3\left( {x - 5} \right)}}{{3 - x}} = 0\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to 5} \dfrac{{x + 3}}{{3 - x}} = - 4\).
Chọn A