LT1

Ta có:

\(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2} = 0\).

\(\mathop {\lim }\limits_{x \to {0^-}} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} {(-x)} = 0\).

Suy ra,\(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right)\).

Vậy hàm số liên tục tại 0.