Cho hàm số \(f\left( x \right) = \dfrac{{2{x^2} + x - 3}}{{x - 1}}\). Dùng định nghĩa tính \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).
Giải chi tiết:
Ta có:
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - 5} \right] = \mathop {\lim }\limits_{x \to 1} \left[ {\dfrac{{2{x^2} + x - 3}}{{x - 1}} - 5} \right]\\ = \mathop {\lim }\limits_{x \to 1} \left[ {\dfrac{{2{x^2} + x - 3 - 5x + 5}}{{x - 1}}} \right] = \mathop {\lim }\limits_{x \to 1} \left[ {\dfrac{{2{x^2} - 4x + 2}}{{x - 1}}} \right]\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{2{{\left( {x - 1} \right)}^2}}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} 2\left( {x - 1} \right) = 2\left( {1 - 1} \right) = 0\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 5\).
Chọn B.