Cho hàm số f ( x ) = { (x^2 − 4)/( x − 2) khi x > 2 ; ax + 2024 khi x ≤ 2 .

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Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{{x^2} - 4}}{{x - 2}}\;\;\;\;\;\;{\rm{khi}}\;x > 2\\ax + 2024\;{\rm{khi}}\;x \le 2\end{array} \right.\).

\(f\left( 2 \right) = 0\).

\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 4\).

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = - 4\).

\(a = - 1010\) thì tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

Giải thích

a) \(f\left( 2 \right) = 2a + 2024\).

b) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 4}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( {x + 2} \right) = 4\).

c) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {ax + 2024} \right) = 2a + 2024\).

d) Để tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) thì \(2a + 2024 = 4 \Leftrightarrow a = - 1010\).

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