Tìm khẳng định sai?

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PHẦN I. TRẮC NGHIỆM NHIỀU LỰA CHỌN

Cho \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 3\). Tìm khẳng định sai?

\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) + 3} \right] = 6\).

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

\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 2x} \right] = - 1\).

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

Giải thích

\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) + 3} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) + 3 = 3 + 3 = 6\).

\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 2x} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - \mathop {\lim }\limits_{x \to 2} 2x = 3 - 4 = - 1\).

\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - {x^2}} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - \mathop {\lim }\limits_{x \to 2} {x^2} = 3 - 4 = - 1\).

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