Cho hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\dfrac{{2{x^2} - 5x + 3}}{{x - 1}}\,\,khi\,\,x > 1}\\{\,\,\,\,3x - 4\,\,\,\,\,\,\,\,\,\,khi\,\,x \le 1}\end{array}} \right.\). Tìm \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right),\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) nếu có.
A.\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 1} f\left( x \right)=-1\)
B.\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 1} f\left( x \right)=-\dfrac{5}{3}\)
C.\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 1} f\left( x \right)= -\infty\)
D.Không tồn tại.
A.\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 1} f\left( x \right)=-1\)
B.\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 1} f\left( x \right)=-\dfrac{5}{3}\)
C.\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 1} f\left( x \right)= -\infty\)
D.Không tồn tại.
Loga
Hóa Học lớp 12
