a) Khi m = −1 thì lim x → 2 − f ( x ) = 1 .

13/20

Cho hàm số\(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 1 + 2m\;khi\;\;x < 2\\\sqrt {x + 7} \;\;\;\;\;\;\;\;khi\;\;x \ge 2\end{array} \right.\)(m là tham số).

a) Khi m = −1 thì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 1\).

b) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = 5\).

c) Tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) khi m = −3.

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

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

a) Với m = −1 thì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 1 - 2} \right) = 4 - 1 - 2 = 1\).

b) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \sqrt {x + 7} = \sqrt {3 + 7} = \sqrt {10} \).

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

\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 7} = 3\).

Để tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) thì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\) Û 3 + 2m = 3 Û m = 0.

d) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 7} = 3\).

Đáp án: a) Đúng; b) Sai; c) Sai; d) Đúng.