Giải thích các bước giải:
\(\begin{array}{l}
b,\\
\mathop {\lim }\limits_{x \to 3} \frac{{2 - \sqrt[3]{{x + 1}}}}{{3x - 9}}\\
\mathop {\lim }\limits_{x \to 3} \left( {2 - \sqrt[3]{{x + 1}}} \right) = 2 - \sqrt[3]{{3 + 1}} = 2 - \sqrt[3]{4}\\
\mathop {\lim }\limits_{x \to 3} \left( {3x - 9} \right) = 3.3 - 9 = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to 3} \frac{{2 - \sqrt[3]{{x + 1}}}}{{3x - 9}} = \infty \\
c,\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{{{\left( {x - 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x - 2}}\\
\mathop {\lim }\limits_{x \to 2} \left( {x - 1} \right) = 2 - 1 = 1\\
\mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right) = 2 - 2 = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x - 2}} = \infty \\
d,\\
\mathop {\lim }\limits_{x \to 1} \frac{{4{x^6} - 5{x^5} + x}}{{{{\left( {1 - x} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {4{x^5} - 5{x^4} + 1} \right)}}{{{{\left( {1 - x} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x\left[ {\left( {4{x^5} - 4{x^4}} \right) - \left( {{x^4} - 1} \right)} \right]}}{{{{\left( {1 - x} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x.\left[ {4{x^4}\left( {x - 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)} \right]}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x.\left[ {\left( {x - 1} \right).\left( {4{x^4} - \left( {x + 1} \right)\left( {{x^2} + 1} \right)} \right)} \right]}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x.\left( {4{x^4} - {x^3} - {x^2} - x - 1} \right)}}{{\left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x.\left( {4{x^4} - 4{x^3} + 3{x^3} - 3{x^2} + 2x - 2x + x - 1} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {x - 1} \right)\left( {4{x^3} + 3{x^2} + 2x + 1} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} x\left( {4{x^3} + 3{x^2} + 2x + 1} \right)\\
= 1.\left( {{{4.1}^3} + {{3.1}^2} + 2.1 + 1} \right)\\
= 10
\end{array}\)
