Bài tập 5.9 trang 199 SBT Toán 11

a) Với x ≥ 0 ta có y = x²

\({{\rm{\Delta }}y = {{\left( {x + {\rm{\Delta }}x} \right)}^2} - {x^2} = 2x{\rm{\Delta }}x + {{\left( {{\rm{\Delta }}x} \right)}^2} \Rightarrow \frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}} = 2x + {\rm{\Delta }}x}\)
Vậy \(\mathop {\lim }\limits_{{\rm{\Delta }}x \to {0^ + }} \frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}} = \mathop {\lim }\limits_{{\rm{\Delta }}x \to {0^ + }} \left( {2x + {\rm{\Delta }}x} \right) = 2x\)
Do vậy \(\mathop {\lim }\limits_{{\rm{\Delta }}x \to {0^ + }} \frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}}\) tại x = 0 là 0   
b) 
a) Với x < 0 ta có y = x
\({{\rm{\Delta }}y = \left( {x + {\rm{\Delta }}x} \right) - x = {\rm{\Delta }}x \Rightarrow \frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}} = 1}\)
Vậy \(\mathop {\lim }\limits_{{\rm{\Delta }}x \to {0^ + }} \frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}} = 1\)
Do vậy \(\mathop {\lim }\limits_{{\rm{\Delta }}x \to {0^ - }} \frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}}\) tại x = 0 là 1
Chọn D.