Giải giúp vs ạ 😁😶🙂🙂🙂

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
*)\\
\mathop {\lim }\limits_{x \to 0} \frac{{x + 1 + \sqrt[3]{{x - 1}}}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{x}{x} + \frac{{1 + \sqrt[3]{{x - 1}}}}{x}} \right]\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{\left( {1 + \sqrt[3]{{x - 1}}} \right)\left( {{1^2} - 1.\sqrt[3]{{x - 1}} + {{\sqrt[3]{{x - 1}}}^2}} \right)}}{{x.\left( {{1^2} - 1.\sqrt[3]{{x - 1}} + {{\sqrt[3]{{x - 1}}}^2}} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{{1^3} + \left( {x - 1} \right)}}{{x.\left( {1 - \sqrt[3]{{x - 1}} + {{\sqrt[3]{{x - 1}}}^2}} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{x}{{x.\left( {1 - \sqrt[3]{{x - 1}} + {{\sqrt[3]{{x - 1}}}^2}} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{1}{{1 - \sqrt[3]{{x - 1}} + {{\sqrt[3]{{x - 1}}}^2}}}} \right]\\
 = 1 + \frac{1}{{1 - \sqrt[3]{{0 - 1}} + {{\sqrt[3]{{0 - 1}}}^2}}} = 1 + \frac{1}{3} = \frac{4}{3}\\
*)\\
\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - {x^2} + 2x - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2}\left( {x - 1} \right) + 2\left( {x - 1} \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + 2} \right) = {1^2} + 2 = 3
\end{array}\)