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Câu hỏi: Cho hàm số
\(f\left( x \right) = \left\{ {\matrix{{{x^2} - 2x + 3 \text{ với } x \le 2.} \cr {4x - 3 \text{ với } x > 2} \cr} } \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right),\mathop {\lim }\limits_{x \to {2^ - }} f\left(x \right) \text{ và } \mathop {\lim }\limits_{x \to 2} f\left(x \right)\) (nếu có).
\(f\left( x \right) = \left\{ {\matrix{{{x^2} - 2x + 3 \text{ với } x \le 2.} \cr {4x - 3 \text{ với } x > 2} \cr} } \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right),\mathop {\lim }\limits_{x \to {2^ - }} f\left(x \right) \text{ và } \mathop {\lim }\limits_{x \to 2} f\left(x \right)\) (nếu có).
Lời giải chi tiết
Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} f\left(x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left({4x - 3} \right) =4.2-3= 5 \cr
& \mathop {\lim }\limits_{x \to {2^ - }} f\left(x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left({{x^2} - 2x + 3} \right) =2^2-2.2+3= 3 \cr} \)
Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left(x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)
Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} f\left(x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left({4x - 3} \right) =4.2-3= 5 \cr
& \mathop {\lim }\limits_{x \to {2^ - }} f\left(x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left({{x^2} - 2x + 3} \right) =2^2-2.2+3= 3 \cr} \)
Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left(x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)