Đáp án:
\[\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^3} - 1}}{{\sqrt {{x^2} - 1} }} = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^3} - 1}}{{\sqrt {{x^2} - 1} }}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\sqrt {\left( {x - 1} \right)\left( {x + 1} \right)} }}\\
= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\sqrt {x - 1} .\sqrt {x + 1} }}\,\,\,\,\,\,\,\,\,\left( {x \to {1^ + } \Rightarrow x > 1 \Rightarrow \left\{ \begin{array}{l}
x - 1 > 0\\
x + 1 > 0
\end{array} \right.} \right)\\
= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {x - 1} \left( {{x^2} + x + 1} \right)}}{{\sqrt {x + 1} }}\\
= \frac{{\sqrt {1 - 1} .\left( {{1^2} + 1 + 1} \right)}}{{\sqrt {1 + 1} }}\\
= 0
\end{array}\)
