Cho hàm số \(f(x) = \left\{ \begin{array}{l}{x^2} - 3{\,\,\rm{ khi\,\, x}} \ge {\rm{2}}\\x - 1{\,\,\rm{ khi \,\,x < 2}}\end{array} \right.\). Chọn kết quả đúng của \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).
-
A.
0
-
B.
1
-
C.
2
-
D.
3
\(\mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}^ - } f\left( x \right) = f\left( {{x_0}} \right)\).
Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} - 3} \right) = {2^2} - 3 = 1\\\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 1} \right) = 2 - 1 = 1\end{array}\)
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} f\left( x \right) = 1\).
Đáp án : B
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