\(\mathop {\lim \to {1^ + }} {{x^3} - {x^2}} }}{{\sqrt {x - 1} + 1 - x}}\) bằng:

* Hướng dẫn giải

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^3} - {x^2}} }}{{\sqrt {x - 1}  + 1 - x}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^2}\left( {x - 1} \right)} }}{{\sqrt {x - 1}  - \sqrt {{{\left( {x - 1} \right)}^2}} }}\\
 = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {x - 1} }}{{\sqrt {x - 1} \left( {1 - \sqrt {x - 1} } \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{x}{{1 - \sqrt {x - 1} }} = 1
\end{array}\)