Hoạt động khám phá 1 trang 80 SGK Toán 11 Chân trời sáng tạo tập 1 - CTST

Phương pháp giải:

Bước 1: Tính các giới hạn một bên \(\mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right),\mathop {\lim }\limits_{x \to {x_0}^ - } {\rm{ }}f\left( x \right)\).

Bước 2: So sánh \(\mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right),\mathop {\lim }\limits_{x \to {x_0}^ - } {\rm{ }}f\left( x \right)\)

• Nếu \(\mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right) \)\( = \mathop {\lim }\limits_{x \to {x_0}^ - } {\rm{ }}f\left( x \right) = L\) thì \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L\).

• Nếu \(\mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right)\)\(  \ne \mathop {\lim }\limits_{x \to {x_0}^ - } {\rm{ }}f\left( x \right)\) thì không tồn tại \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right)\).

Lời giải chi tiết:

• \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\)\(  = \mathop {\lim }\limits_{x \to {1^ + }} \left( {1 + x} \right) = 1 + 1 = 2\).

\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \)\( = \mathop {\lim }\limits_{x \to {1^ - }} 1 = 1\).

Vì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \)\( \ne \mathop {\lim }\limits_{x \to {1^ - }} {\rm{ }}f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).

• \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \)\( = \mathop {\lim }\limits_{x \to {2^ + }} \left( {5 - x} \right) \)\( = 5 - 2 = 3\).

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) \)\( = \mathop {\lim }\limits_{x \to {2^ - }} \left( {1 + x} \right) \)\( = 1 + 2 = 3\).

Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \)\( = \mathop {\lim }\limits_{x \to {2^ - }} {\rm{ }}f\left( x \right) = 3\) nên \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 3\).

Ta có: \(f\left( 2 \right) \)\( = 1 + 2 = 3\).

Vậy \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)\(  = f\left( 2 \right)\).

-- Mod Toán 11 HỌC247