Đáp án:
15) \(\dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
13)\mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{2 - x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right)\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{1 - x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 1}}{{x + 1}}\\
= \dfrac{{ - 1}}{2}\\
14)\mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{1 + x + {x^2} - 3}}{{\left( {1 - x} \right)\left( {{x^2} + x + 1} \right)}}} \right)\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {1 - x} \right)\left( {{x^2} + x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - x - 2}}{{{x^2} + x + 1}}\\
= \dfrac{{ - 1 - 2}}{{1 - 1 + 1}} = - 3\\
15)\mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left( {\sqrt[3]{x} - 1} \right)}^2}}}{{{{\left( {\sqrt[3]{{{x^3}}} - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left( {\sqrt[3]{x} - 1} \right)}^2}}}{{{{\left[ {\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)} \right]}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}}\\
= \dfrac{1}{{1 + 1 + 1}} = \dfrac{1}{3}
\end{array}\)
